Curve Sketching | NEB Class 11 Mathematics Notes

Chapter Overview

Curve sketching is the art and science of visualizing mathematical functions on a coordinate plane. While an equation tells you the what, a graph shows you the how—revealing patterns, trends, and behavior that would remain hidden in algebraic form alone. Imagine a doctor reading an ECG (electrocardiogram): the graph of heart voltage over time tells more than any number could. In this chapter, we transform algebraic expressions into visual representations by systematically analyzing function properties and applying transformations.

What this chapter covers

Even and Odd Functions: Definition and Graphical SymmetryPeriodic Functions and Finding Fundamental PeriodsSymmetry Tests: About x-axis, y-axis, and OriginMonotonicity: Increasing and Decreasing BehaviorAsymptotes: Vertical and Horizontal ApproachesFunction Transformations: Shifts, Stretches, and ReflectionsIntercepts: Finding x and y intersection pointsSketching Elementary Functions: Linear, Quadratic, Polynomial, Rational, Exponential, Logarithmic, and TrigonometricDomain and Range AnalysisVertex, Holes, and Discontinuities in Graphs

Learning Objectives

1

Identify and prove whether a function is even, odd, or neither using algebraic tests and graphical intuition.

2

Determine the periodicity of functions and calculate fundamental periods using the formula $P = \frac{2\pi}{m}$ for trigonometric functions.

3

Test curves for symmetry about the x-axis, y-axis, and origin using substitution methods.

4

Analyze monotonicity by determining where functions are increasing or decreasing across their domains.

5

Identify and sketch vertical and horizontal asymptotes for rational, exponential, and logarithmic functions.

6

Apply transformations (horizontal shifts, vertical shifts, reflections, stretches) to create new graphs from basic functions.

7

Find x-intercepts (zeros) and y-intercepts by setting appropriate variables to zero.

8

Sketch complete, accurate graphs of polynomial, rational, exponential, logarithmic, and trigonometric functions.

9

Calculate and interpret domain and range for complex functions with multiple restrictions.

10

Recognize and label special features including vertices, holes, and discontinuities on graphs.

The Five-Step Sketching Framework

Key Insight

Every graph, no matter how complex, can be sketched systematically by following five key steps. This framework transforms curve sketching from an overwhelming task into a manageable process.

The Five Steps Are:

1. Find the Domain: Identify all values of $x$ for which the function is defined. Exclude where denominators are zero, where roots contain negatives, and where logarithm arguments are non-positive.

2. Find Intercepts: Set $x = 0$ for y-intercepts and $y = 0$ for x-intercepts. These points anchor the graph and provide crucial reference marks.

3. Test for Symmetry: Check if the function is even ($f(-x) = f(x)$), odd ($f(-x) = -f(x)$), or periodic. This halves or thirds your work by revealing pattern repetition.

4. Find Asymptotes and Behavior: Identify vertical asymptotes (where denominator = 0), horizontal asymptotes (as $x \to \pm\infty$), and overall end behavior.

5. Plot Key Points and Connect: Use critical points (vertices, maxima, minima) and the information above to sketch the complete curve accurately.

This framework is not just helpful—it is essential for NEB exams. Examiners award marks for showing ALL steps, not just for the final drawing. Missing a step may cost 2-3 marks even if your graph looks reasonable.

Key Definitions

Even Function

A function $f(x)$ is even if and only if $f(-x) = f(x)$ for all $x$ in its domain. Graphically, an even function is symmetric about the $y$-axis. This means if you fold the graph along the $y$-axis, both halves match perfectly. Common examples: $f(x) = x^2$, $f(x) = \cos x$, $f(x) = |x|$.

Odd Function

A function $f(x)$ is odd if and only if $f(-x) = -f(x)$ for all $x$ in its domain. Graphically, an odd function is symmetric about the origin. If you rotate the graph 180° about the origin, it looks identical. Common examples: $f(x) = x^3$, $f(x) = \sin x$, $f(x) = \frac{1}{x}$.

Periodic Function

A function $f(x)$ is periodic with period $p$ if $f(x + p) = f(x)$ for all $x$ in the domain, where $p$ is the smallest positive number satisfying this property. The period represents the horizontal distance after which the function's pattern repeats exactly. For example, $\sin x$ has period $2\pi$, and $\tan x$ has period $\pi$.

Vertical Asymptote

A vertical line $x = a$ is a vertical asymptote of a function if as $x$ approaches $a$, the function value approaches positive or negative infinity. Graphically, the curve gets arbitrarily close to the line $x = a$ but never touches it at a finite point. Vertical asymptotes occur where denominators equal zero (in rational functions) or where logarithmic arguments approach zero.

Horizontal Asymptote

A horizontal line $y = b$ is a horizontal asymptote of a function if as $x$ approaches positive or negative infinity, the function value approaches $b$. Mathematically, $\lim_{x \to \infty} f(x) = b$ or $\lim_{x \to -\infty} f(x) = b$. For rational functions, this depends on the degrees of numerator and denominator polynomials.

Monotonic Function

A function is monotonically increasing on an interval if whenever $x_1 < x_2$, we have $f(x_1) < f(x_2)$. Similarly, a function is monotonically decreasing on an interval if whenever $x_1 < x_2$, we have $f(x_1) > f(x_2)$. These properties describe the 'direction' a graph travels across an interval.

Domain of a Function

The domain is the set of all possible input values ($x$-values) for which the function is defined. For example, the domain of $f(x) = \sqrt{x}$ is $[0, \infty)$ because we cannot take square roots of negative numbers (in real numbers). The domain is restricted by denominators, roots, and logarithmic arguments.

Range of a Function

The range is the set of all possible output values ($y$-values) that the function actually produces. While the co-domain is the 'target set', the range is what is actually achieved. For example, the range of $f(x) = x^2$ is $[0, \infty)$, not $\mathbb{R}$, because squares are never negative.

Vertex of a Parabola

The vertex is the turning point (peak or valley) of a quadratic function $f(x) = ax^2 + bx + c$. For a parabola $y = ax^2 + bx + c$, the x-coordinate of the vertex is $x = -\frac{b}{2a}$, and the y-coordinate is found by substituting this $x$ value back into the function. The vertex lies on the axis of symmetry of the parabola.

Function Transformation

A transformation modifies a function's graph without changing its fundamental shape. Common transformations include: (1) Horizontal shift: $f(x - h)$ shifts right by $h$; (2) Vertical shift: $f(x) + k$ shifts up by $k$; (3) Reflection: $-f(x)$ reflects across x-axis; $f(-x)$ reflects across y-axis; (4) Stretching: $af(x)$ stretches vertically by factor $|a|$.

Even vs. Odd Functions: The Complete Comparison

Even and odd functions are fundamental to understanding symmetry in mathematics. This comparison table helps you quickly distinguish between them and apply the correct graphing techniques.

Even Function

$$$f(-x) = f(x)$$$

Symmetry: Mirror image about the y-axis. If point $(a, b)$ is on the graph, then $(-a, b)$ is also on the graph.

Algebraic Test: Replace $x$ with $-x$. If you get the original function back, it is even.

Examples: $f(x) = x^2$, $f(x) = \cos x$, $f(x) = |x|$, $f(x) = x^4 + 1$

Graphing Advantage: Sketch only the right side ($x \geq 0$) and mirror it to the left side.

Domain Requirement: The domain must be symmetric about zero (e.g., $[-5, 5]$ or $\mathbb{R}$, not $[0, 10]$).

VS

Odd Function

$$$f(-x) = -f(x)$$$

Symmetry: 180° rotational symmetry about the origin. If point $(a, b)$ is on the graph, then $(-a, -b)$ is also on the graph.

Algebraic Test: Replace $x$ with $-x$. If you get the negative of the original function, it is odd.

Examples: $f(x) = x^3$, $f(x) = \sin x$, $f(x) = \frac{1}{x}$, $f(x) = x^5 + x$

Graphing Advantage: Sketch only the first quadrant and rotate 180° about the origin for the complete graph.

Special Property: All odd functions pass through the origin: $f(0) = 0$ (because $f(-0) = -f(0)$ means $f(0) = -f(0)$, so $f(0) = 0$).

Quick Test Memory: 'Even = Mirror' (y-axis mirror), 'Odd = Opposite' (opposite sign under negative input). A function is neither even nor odd if it satisfies neither test. No function can be both even and odd except $f(x) = 0$.

Formula Sheet & Key Properties

Period of Sine and Cosine Functions

P = \frac{2\pi}{|m|} \text{ for } f(x) = \sin(mx) \text{ or } f(x) = \cos(mx)

The period is the horizontal distance after which the function repeats. A larger $|m|$ means a shorter period (more compressed).

1By definition of period: $f(x + P) = f(x)$ for all $x$.

2For $f(x) = \sin(mx)$: $\sin(m(x+P)) = \sin(mx)$

3$\sin(mx + mP) = \sin(mx)$

4The sine function repeats when its argument increases by $2\pi$. So: $mP = 2\pi$

5Solving for $P$: $P = \frac{2\pi}{m}$

Period of Tangent Function

P = \frac{\pi}{|m|} \text{ for } f(x) = \tan(mx)

Tangent has period $\pi$, not $2\pi$. This is because $\tan(x + \pi) = \tan x$.

1By definition: $f(x + P) = f(x)$

2For $f(x) = \tan(mx)$: $\tan(m(x+P)) = \tan(mx)$

3Tangent repeats when its argument increases by $\pi$. So: $mP = \pi$

4Solving: $P = \frac{\pi}{m}$

Vertex of a Quadratic Parabola

\text{Vertex} = \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right) \text{ for } f(x) = ax^2 + bx + c

The x-coordinate of the vertex is $-\frac{b}{2a}$. Substitute this into the function to get the y-coordinate.

1Completing the square for $f(x) = ax^2 + bx + c$:

2$f(x) = a\left(x^2 + \frac{b}{a}x\right) + c$

3$= a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right) - \frac{b^2}{4a} + c$

4$= a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}$

5The vertex form shows the vertex at $x = -\frac{b}{2a}$ with $y = c - \frac{b^2}{4a}$

Horizontal Asymptote for Rational Functions

\text{If degree of numerator} < \text{degree of denominator, then } y = 0 \text{ is the asymptote}

Compare the degrees of the polynomials in the numerator and denominator to determine horizontal behavior.

1For $f(x) = \frac{a_nx^n + ...}{b_mx^m + ...}$:

2If $n < m$: as $x \to \pm\infty$, the denominator grows faster, so $f(x) \to 0$

3If $n = m$: $f(x) \to \frac{a_n}{b_m}$ (ratio of leading coefficients)

4If $n > m$: no horizontal asymptote (function grows without bound)

Transformation Rules

y = a \cdot f(b(x - h)) + k

This master formula encompasses all transformations: shifts, stretches, and reflections.

1Horizontal Shift: Replace $x$ with $(x - h)$ to shift right by $h$ units (left if $h$ is negative)

2Vertical Shift: Add $k$ to shift up by $k$ units (down if $k$ is negative)

3Vertical Stretch/Compression: Multiply by $a$ to stretch (if $|a| > 1$) or compress (if $|a| < 1$)

4Horizontal Stretch/Compression: Replace $x$ with $bx$ to compress (if $|b| > 1$) or stretch (if $|b| < 1$)

5Reflection Across x-axis: Negate the entire function: $y = -f(x)$

6Reflection Across y-axis: Replace $x$ with $-x$: $y = f(-x)$

Domain Restrictions Summary

\text{Denominators} \neq 0, \quad \text{Under even roots} \geq 0, \quad \text{Logarithm arguments} > 0

These three rules cover 95% of domain restriction problems in NEB exams.

1Fractions: The denominator of any fraction cannot be zero. If $g(x)$ is in a denominator, solve $g(x) = 0$ and exclude those values.

2Even Roots: The expression under a square root (or any even root) must be non-negative. Set the radicand $\geq 0$ and solve.

3Logarithms: The argument of any logarithm must be strictly positive (not just non-negative). If $\log(h(x))$ appears, solve $h(x) > 0$.

Step-by-Step: The Complete Curve Sketching Process

Follow this logical sequence to sketch any function accurately. Examiners reward students who show all steps, not just the final sketch.

Step 1: Can you identify the Domain?

Yes

Check for denominators = 0, negative values under even roots, and logarithm arguments ≤ 0. Write the domain explicitly (e.g., Domain = $\mathbb{R} - \{2\}$ or $[0, \infty)$).

No

This is the most common step skipped. Review domain restrictions above.

Step 2: Can you find the Intercepts?

Yes

For y-intercept, set $x = 0$ and solve. For x-intercepts, set $y = 0$ and solve. Plot these points on your sketch.

No

Continue to next step. Some functions may have no intercepts in the domain.

Step 3: Test for Symmetry (Even/Odd/Periodic)?

Yes

Test: $f(-x) = f(x)$ (even), $f(-x) = -f(x)$ (odd), or $f(x+p) = f(x)$ (period $p$). This halves or thirds your work!

No

Proceed without symmetry advantage. You must sketch the entire graph.

Step 4: Are there Asymptotes (Vertical/Horizontal)?

Yes

Vertical asymptotes: set denominators = 0. Horizontal asymptotes: analyze limits as $x \to \pm\infty$. Draw these lines as dashed guides.

No

Some functions (polynomial, exponential) have no asymptotes. Continue.

Step 5: Can you find Special Points (Vertex, Turning Points)?

Yes

For quadratics, use vertex formula $x = -b/2a$. For other functions, use calculus (if learned) or test critical values. Mark these points.

No

Use general behavior and the points you already have.

Step 6: Determine End Behavior (what happens as $x \to \pm\infty$)?

Yes

Check the degree and sign of leading coefficient (polynomial), or use limits. This guides your sketch endpoints.

No

For periodic functions, the pattern repeats. For bounded functions, behavior is limited.

Step 7: Connect Points and Sketch Smoothly.

Yes

Use all information above to draw a smooth, accurate curve. Show asymptotes as guides. Use arrows to indicate direction and end behavior.

No

You are done with analysis. Now execute the sketch.

Top 7 Memory Tricks for Curve Sketching Success

Trick: These mnemonics and shortcuts will save you time and prevent common errors during exams.

Steps to Remember:

Trick 1 - 'EO' for Even-Odd: Even = Equal ($f(-x) = f(x)$, like $x^2$). Odd = Opposite ($f(-x) = -f(x)$, like $x^3$). This rhyme helps you remember which is which instantly.
Trick 2 - 'Y-Axis Mirror' vs 'Origin Spin': Even functions look like a mirror image reflected across the y-axis. Odd functions look like the graph has been spun 180° around the origin. Visualize this instead of memorizing definitions.
Trick 3 - 'Shift Paradox' (The Plus-Minus Swap): This confuses many students! In $f(x+h)$, the $+$ means shift LEFT (not right). In $f(x-h)$, the $-$ means shift RIGHT (not left). Think 'opposite direction'. Example: $(x+2)^2$ shifts the parabola 2 units LEFT of origin.
Trick 4 - 'SIN-cos: 2π Period; TAN: π Period': Remember $\sin(x)$ and $\cos(x)$ repeat every $2\pi$ (one full revolution). $\tan(x)$ repeats every $\pi$ (half revolution) because of its discontinuities. Use $P = \frac{2\pi}{m}$ for sine/cosine and $P = \frac{\pi}{m}$ for tangent.
Trick 5 - 'Asymptotes are Approaching, Not Reaching': The curve gets arbitrarily close to an asymptote but never actually touches it (in the limit). If you see intersection points, it may be a hole, not an asymptote. A vertical asymptote represents an infinite jump; the function is undefined there.
Trick 6 - 'Domain has Three Enemies': The three things that restrict domain are: (1) Division by Zero, (2) Square Root of Negative, (3) Log of Non-Positive. Always check for these three before claiming a domain is all real numbers.
Trick 7 - 'Vertex is on the Axis of Symmetry': For $f(x) = ax^2 + bx + c$, the vertex $x = -b/2a$ lies on the axis of symmetry $x = -b/2a$. This formula never changes. The parabola is symmetric about this vertical line.

Mnemonic: Remember the 5-Step Framework: Domain → Intercepts → Symmetry → Asymptotes → Points → Sketch. This order is not random; each step depends on the previous ones.

Common Mistakes in NEB Curve Sketching Exams

These errors appear in 80% of incomplete student answers. Study them carefully to avoid losing marks.

Mistake 1 - Confusing Period of tan(x) with sin(x): Students write: 'The period of $\tan x$ is $2\pi$.' WRONG! $\tan x$ has period $\pi$, not $2\pi$. This is because $\tan(x + \pi) = \tan x$. Since $\tan x$ has vertical asymptotes at $x = \pm\pi/2$, the pattern repeats faster than sine/cosine. This single mistake costs 2-3 marks when sketching trigonometric functions.
Mistake 2 - Forgetting the Negative Sign in Vertex Formula: A student computes the vertex of $y = x^2 - 4x + 3$ as: 'x = 4/2 = 2'. They forgot the minus! The correct formula is $x = -b/2a = -(-4)/2(1) = 2$. The negative sign is non-negotiable. Verification: $f(2) = 4 - 8 + 3 = -1$, so vertex is $(2, -1)$.
Mistake 3 - Sketching Logarithmic Functions for Negative x: A student draws the curve $y = \log x$ extending into the second quadrant (negative x). Impossible! The domain of any logarithm is strictly positive: $x > 0$. The graph exists only to the right of the y-axis. The y-axis ($x = 0$) is a vertical asymptote.
Mistake 4 - Not Distinguishing Vertical vs Horizontal Asymptotes: A student writes: 'The asymptote of $y = 1/(x-3)$ is $x = 0$.' Incomplete! There are TWO asymptotes: (1) Vertical: $x = 3$ (denominator = 0), (2) Horizontal: $y = 0$ (as $x \to \infty$). Each tells a different story about the function's behavior.
Mistake 5 - Missing Domain Restrictions with Multiple Conditions: Given $f(x) = \sqrt{x}/(x-2)$, a student writes: 'Domain = $x \geq 0$'. Incomplete! You must exclude $x = 2$ (denominator). Correct domain: $[0, 2) \cup (2, \infty)$. When multiple restrictions exist, combine them using set notation.
Mistake 6 - Confusing Amplitude with Period: The function $y = 3\sin(2x)$ has amplitude 3 (vertical stretch) and period $\pi$ (from $2\pi/2$). A student writes: 'Period = 3'. The amplitude is 3, but period is $\pi$! These are completely different properties.
Mistake 7 - Holes vs Asymptotes (Rational Functions): Given $y = (x-2)/(x-2)$, a student writes: 'Vertical asymptote at $x = 2$'. WRONG! The function simplifies to $y = 1$ (for $x \neq 2$). There is a hole (a point removed from the graph) at $(2, 1)$, not an asymptote. Learn to factor and cancel common terms before identifying asymptotes.
Mistake 8 - Treating Even/Odd as Binary (Either/Or): Some functions are neither even nor odd. A student checks $f(-x) = f(x)$, finds it false, then automatically concludes the function is odd. Always verify BOTH conditions. Test: $f(x) = 3x + 4$. Is $f(-x) = 3(-x) + 4 = -3x + 4 = f(x)$? No. Is $f(-x) = -f(x)$? Is $-3x + 4 = -(3x+4) = -3x - 4$? No. Therefore, it is neither even nor odd.

Solved Examples (Easy → Medium → Hard)

1

Determine if $f(x) = x^2 + 3$ is even, odd, or neither.

Easy

Solution

To determine if a function is even or odd, we apply the algebraic tests by replacing $x$ with $-x$.

Given: $f(x) = x^2 + 3$

Step 1: Calculate $f(-x)$

Replace every $x$ in the function with $-x$:
$$f(-x) = (-x)^2 + 3$$

Step 2: Simplify

$$f(-x) = x^2 + 3$$

Step 3: Compare with $f(x)$ and $-f(x)$

We have:
- $f(-x) = x^2 + 3$
- $f(x) = x^2 + 3$
- $-f(x) = -(x^2 + 3) = -x^2 - 3$

Since $f(-x) = f(x)$, the function is EVEN.

Graphical Interpretation: This parabola is symmetric about the y-axis. If you fold the graph along the y-axis, both halves match perfectly.

Graph Description:

y | 5 + • • | / \ / \ 4 + / \ / \ | / \ / \ 3 + •-------•---•-------• | \ / 2 + • | 1 + | 0 +----------+----------+-------> x -2 -1 0 1 2

The parabola opens upward with vertex at $(0, 3)$. The dashed line $x = 0$ (the y-axis) is the axis of symmetry. The left half mirrors the right half perfectly.

∴ The function is EVEN because $f(-x) = f(x)$.

2

Find the period of $f(x) = \sin(3x)$.

Easy

Solution

To find the period of a sine function, we use the formula for trigonometric periodicity.

Given: $f(x) = \sin(3x)$

The general form is $f(x) = \sin(mx)$, where $m$ is the coefficient of $x$.

Step 1: Identify the value of $m$

Comparing $f(x) = \sin(3x)$ with $f(x) = \sin(mx)$, we see that $m = 3$.

Step 2: Apply the period formula for sine

For the function $\sin(mx)$, the period is:
$$P = \frac{2\pi}{m}$$

Step 3: Substitute $m = 3$

$$P = \frac{2\pi}{3}$$

Verification: The standard sine function $\sin(x)$ has period $2\pi$. Since we have $\sin(3x)$, the function oscillates 3 times faster, so the period is $1/3$ of the original: $2\pi / 3$ ✓

Graphical Meaning: The pattern of the sine wave repeats every $\frac{2\pi}{3}$ units along the x-axis.

Graph Description:


        y = sin(3x)
        |
      1 +   •       •       •
        |  / \     / \     / \
      0 +—•———•———•———•———•———• x
   0   π/6 π/3 π/2 2π/3 5π/6 π
        |       \     /       \
     -1 +   •       •       •
        |
     Period = 2π/3

The sine curve completes one full oscillation (zero → max → zero → min → zero) in the span of $2\pi/3$ on the x-axis.

∴ The period is $P = \frac{2\pi}{3}$.

3

Find the vertex of the parabola $y = x^2 - 4x + 3$.

Easy

Solution

The vertex is the turning point of a parabola. We use the vertex formula to find it.

Given: $y = x^2 - 4x + 3$

This is in the form $y = ax^2 + bx + c$ where $a = 1$, $b = -4$, $c = 3$.

Step 1: Find the x-coordinate of the vertex using the formula

The x-coordinate of the vertex is:
$$x = -\frac{b}{2a}$$

Substitute the values:
$$x = -\frac{(-4)}{2(1)} = \frac{4}{2} = 2$$

Step 2: Find the y-coordinate by substituting $x = 2$ into the function

$$y = (2)^2 - 4(2) + 3$$
$$y = 4 - 8 + 3$$
$$y = -1$$

Step 3: Write the vertex as an ordered pair

Vertex = $(2, -1)$

Graphical Interpretation: This parabola opens upward (since $a = 1 > 0$), has its minimum point at $(2, -1)$, and is symmetric about the vertical line $x = 2$.

Graph Description:


        y
        |
      3 +   •           •
        |  / \         / \
      2 + /   \       /   \
        |/     \     /     \
      1 +       •   •       •
        |        \ / \
      0 +————————•———•————————x
    0      1    2    3   4
        |        \   /
     -1 +         • (vertex at (2,-1))
        |         |
        Axis of symmetry: x = 2

The parabola has its lowest point (vertex) at $(2, -1)$. It is symmetric about the vertical line $x = 2$. The y-intercept is at $(0, 3)$, and the x-intercepts are at $(1, 0)$ and $(3, 0)$.

∴ The vertex is $(2, -1)$.

4

Test the symmetry of the curve $x = y^2 + 4$.

Medium

Solution

To test for symmetry, we check three conditions: symmetry about the x-axis, y-axis, and origin.

Given: $x = y^2 + 4$

Test 1: Symmetry about the x-axis

Replace $y$ with $-y$ and see if the equation remains unchanged:
$$x = (-y)^2 + 4$$
$$x = y^2 + 4$$ ✓

The equation is identical! Therefore, the curve IS symmetric about the x-axis.

Test 2: Symmetry about the y-axis (for completeness)

Replace $x$ with $-x$:
$$-x = y^2 + 4$$
$$x = -y^2 - 4$$

This is NOT the same as the original equation. The curve is NOT symmetric about the y-axis.

Test 3: Symmetry about the origin (for completeness)

Replace both $x$ with $-x$ and $y$ with $-y$:
$$-x = (-y)^2 + 4$$
$$-x = y^2 + 4$$
$$x = -y^2 - 4$$

This is NOT the same as the original. The curve is NOT symmetric about the origin.

Conclusion: The curve $x = y^2 + 4$ is symmetric about the x-axis only. Graphically, this is a parabola opening to the right (since $x$ is isolated), with vertex at $(4, 0)$, and the two halves above and below the x-axis mirror each other.

Graph Description:


        y
        |     upper half
        |     (y > 0)
      2 +   •————————•
        |  /         \
      1 + •           •
        |/             \
      0 +————•————•————— x
        |  4   5   6   7
        |\             /
     -1 + •           •
        |  \         /
     -2 +   •————————•
        |     lower half
        |     (y < 0)
        |
        Axis of symmetry: x-axis (y = 0)
        Vertex at (4, 0)
        Opens to the right

The parabola opens to the right with vertex at $(4, 0)$. The upper half (for $y > 0$) is the mirror image of the lower half (for $y < 0$) about the x-axis.

∴ The curve is symmetric about the x-axis only.

5

Find the domain of $f(x) = \frac{1}{\sqrt{9 - x^2}}$.

Medium

Solution

This function has TWO restrictions: a denominator (cannot be zero) and a square root (radicand must be non-negative).

Given: $f(x) = \frac{1}{\sqrt{9 - x^2}}$

Restriction 1: The expression under the square root must be positive (strictly, since it's in denominator)

For the square root to be defined and non-zero:
$$9 - x^2 > 0$$
$$9 > x^2$$
$$x^2 < 9$$
$$|x| < 3$$
$$-3 < x < 3$$

This gives us the open interval $(-3, 3)$.

Restriction 2: The denominator cannot be zero

The denominator is $\sqrt{9 - x^2}$. We need:
$$\sqrt{9 - x^2} \neq 0$$
$$9 - x^2 \neq 0$$
$$x^2 \neq 9$$
$$x \neq \pm 3$$

So we must exclude $x = 3$ and $x = -3$. (These are already excluded by Restriction 1.)

Step 3: Combine both restrictions

We need $9 - x^2 > 0$ (strictly positive, not just non-negative).

Domain = $(-3, 3)$ (open interval, excluding both endpoints)

In set notation: Domain = $\{x \in \mathbb{R} : -3 < x < 3\}$

Graph Description: The function is defined only between $x = -3$ and $x = 3$. There are vertical asymptotes at both $x = -3$ and $x = 3$, where the function approaches infinity.


        y
        |
        |     undefined     |     undefined
        | (asymptote at     | (asymptote at
        |  x = -3)          |  x = 3)
      4 +              |             |
        |              |             |
      3 +              |             |
        |              |      •      |
      2 +              |    •   •    |
        |              |  •       •  |
      1 +              |•           •|
        |────•────•────•─────•─────•────
      0 + -3    -1.5   0    1.5    3
        |
     -1 +
        |
     Domain: (-3, 3)
     Vertical asymptotes at x = -3 and x = 3

∴ Domain = $(-3, 3)$

6

Identify and locate the asymptotes of $f(x) = \frac{x+1}{x-2}$.

Medium

Solution

This is a rational function. We must find both vertical and horizontal asymptotes.

Given: $f(x) = \frac{x+1}{x-2}$

Step 1: Find the Vertical Asymptote(s)

Vertical asymptotes occur where the denominator equals zero (and the numerator doesn't also equal zero at that point).

Set the denominator equal to zero:
$$x - 2 = 0$$
$$x = 2$$

Check the numerator at $x = 2$: $(2) + 1 = 3 \neq 0$

Since the numerator is non-zero while the denominator is zero, we have:
Vertical Asymptote: $x = 2$

Step 2: Find the Horizontal Asymptote(s)

Compare the degrees of the numerator and denominator polynomials:
- Numerator: $x + 1$ has degree 1
- Denominator: $x - 2$ has degree 1

When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients:
$$y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{1}{1} = 1$$

Horizontal Asymptote: $y = 1$

Verification:
- As $x \to \infty$: $f(x) = \frac{x+1}{x-2} \approx \frac{x}{x} = 1$ ✓
- As $x \to -\infty$: $f(x) \approx 1$ ✓

Graphical Interpretation: The curve has a vertical asymptote at $x = 2$ (the function shoots to $+\infty$ or $-\infty$ as $x$ approaches 2), and approaches the horizontal line $y = 1$ as $x$ goes to either extreme.

Graph Description:


        y
        |          |
      4 +          |       /
        |          |      /
      3 +          |     /
        |          |    /
      2 +          |   / ← approaches y=1
        |          |  / from below
      1 +—————————••—————— (horizontal asymptote)
        |         / \  |
      0 +        /   \ | ← vertical asymptote
        |       /     \|  at x = 2
     -1 +      /       \
        |     /         \
     -2 +    /           \
        |———————|———•———•———|———— x
       -2     0    2    4    6
        |                  \ |
        |                   \|
        |      approaches y=1 from above

∴ Vertical Asymptote: $x = 2$; Horizontal Asymptote: $y = 1$

7

Sketch the graph of $y = (x-1)(x-2)(x-3)$, showing all key features.

Hard

Solution

This is a cubic polynomial. We'll use the five-step sketching framework.

Given: $y = (x-1)(x-2)(x-3)$

Step 1: Find the Domain

This is a polynomial, so Domain = $\mathbb{R}$ (all real numbers).

Step 2: Find the Intercepts

x-intercepts (set $y = 0$):
$$(x-1)(x-2)(x-3) = 0$$
$$x = 1, 2, 3$$

The curve crosses the x-axis at $(1, 0)$, $(2, 0)$, and $(3, 0)$.

y-intercept (set $x = 0$):
$$y = (0-1)(0-2)(0-3) = (-1)(-2)(-3) = -6$$

The curve crosses the y-axis at $(0, -6)$.

Step 3: Test for Symmetry

$$f(-x) = (-x-1)(-x-2)(-x-3) = -(x+1)(x+2)(x+3)$$

This is neither $f(x)$ nor $-f(x)$. No symmetry.

Step 4: Determine the Sign in Each Interval

Test the sign of $y$ in the intervals: $(-\infty, 1)$, $(1, 2)$, $(2, 3)$, $(3, \infty)$

- For $x < 1$ (test $x = 0$): $y = (-1)(-2)(-3) = -6$ (NEGATIVE)
- For $1 < x < 2$ (test $x = 1.5$): $(+)(-)(-)$ = POSITIVE
- For $2 < x < 3$ (test $x = 2.5$): $(+)(+)(-)$ = NEGATIVE
- For $x > 3$ (test $x = 4$): $(+)(+)(+)$ = POSITIVE

Step 5: Analyze End Behavior

Expand the product:
$$y = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$$

The leading term is $x^3$, so:
- As $x \to \infty$: $y \to \infty$ (curve goes up to the right)
- As $x \to -\infty$: $y \to -\infty$ (curve goes down to the left)

Step 6: Find Critical Points (Turning Points)

The cubic has turning points between the zeros. From sign analysis:
- There's a local maximum between $x = 1$ and $x = 2$
- There's a local minimum between $x = 2$ and $x = 3$

Step 7: Sketch the Complete Graph


        y
      4 +
        |     local max
      2 +       •
        |      / \
      0 +—•———•———•———— x
        | 1   2   3
     -2 +|  \ / \n     -4 +   •   •  local min
        |    \ /
     -6 + •(0,-6)
        |

Key Points:
- x-intercepts: (1,0), (2,0), (3,0)
- y-intercept: (0,-6)
- As x→-∞: y→-∞
- As x→+∞: y→+∞
- Local maximum between x=1 and x=2
- Local minimum between x=2 and x=3

Behavior by region:
- x < 1: negative (below x-axis)
- 1 < x < 2: positive (above x-axis) 
- 2 < x < 3: negative (below x-axis)
- x > 3: positive (above x-axis)

The graph shows a 'wave' pattern: starts from lower left (negative region as $x \to -\infty$), crosses x-axis at $x = 1$, rises to a local peak between 1 and 2, crosses x-axis at $x = 2$, dips to a local valley between 2 and 3, crosses x-axis at $x = 3$, and continues upward (positive region as $x \to \infty$).

∴ Cubic curve with x-intercepts at 1, 2, 3; y-intercept at -6; negative for $x < 1$ and $2 < x < 3$; positive for $1 < x < 2$ and $x > 3$; local maximum between x=1 and x=2; local minimum between x=2 and x=3.

8

Describe how to sketch $y = 2^{x-3} + 5$ using transformations of $y = 2^x$.

Hard

Solution

This problem requires understanding function transformations. We start with the basic exponential function and apply transformations step by step.

Given: Transform $y = 2^x$ into $y = 2^{x-3} + 5$

The transformed function can be written as: $y = 2^{(x-3)} + 5$

Step 1: Identify the transformation components

The transformation has the form $y = f(x - h) + k$ where:
- $h = 3$ (horizontal shift)
- $k = 5$ (vertical shift)

Step 2: Apply Horizontal Shift: $y = 2^x \to y = 2^{x-3}$

The expression $(x - 3)$ means shift RIGHT by 3 units (the minus sign indicates rightward movement).

Key points after horizontal shift:
- Original $(0, 1)$ moves to $(3, 1)$ [since $2^{3-3} = 2^0 = 1$]
- Original $(1, 2)$ moves to $(4, 2)$ [since $2^{4-3} = 2^1 = 2$]
- Original $(2, 4)$ moves to $(5, 4)$ [since $2^{5-3} = 2^2 = 4$]
- The horizontal asymptote stays at $y = 0$

Step 3: Apply Vertical Shift: $y = 2^{x-3} \to y = 2^{x-3} + 5$

Adding 5 to the entire function means shift UP by 5 units.

Key points after vertical shift:
- Previous $(3, 1)$ moves to $(3, 6)$
- Previous $(4, 2)$ moves to $(4, 7)$
- Previous $(5, 4)$ moves to $(5, 9)$
- The horizontal asymptote moves from $y = 0$ to $y = 5$

Step 4: Summary of Final Graph Properties

1. Key Point (Anchor): The transformed graph passes through $(3, 6)$ (this corresponds to the original point $(0, 1)$)
2. Shape: Same exponential curve shape as $y = 2^x$ (increasing, concave up)
3. Horizontal Asymptote: $y = 5$ (all y-values approach but never reach 5)
4. Domain: Still $\mathbb{R}$ (all real numbers)
5. Range: $(5, \infty)$ (all values above 5)
6. Monotonicity: Always increasing (as x increases, y increases)

Graph Description:


        y
     10 +                    /
        |                   /
      9 +                 •(5,9)
        |                /
      8 +              /
        |            /
      7 +          •(4,7)
        |        /
      6 +      •(3,6)← Key point (anchor)
        |    /         (moved from (0,1))
      5 +—•———————————— ← horizontal asymptote y=5
        | /  (undefined
      4 + /   values
        |/    below
      3 +     this line)
        |
      2 +
        |
      1 +
        |
      0 +————•————•————•——— x
       -1    1    3    5    7

Transformations Summary:
1. Start with y = 2^x
2. Shift RIGHT 3 units → y = 2^(x-3)
3. Shift UP 5 units → y = 2^(x-3) + 5

Detailed Graph Interpretation:

- At $x = 0$: $y = 2^{0-3} + 5 = 2^{-3} + 5 = \frac{1}{8} + 5 = 5.125$ (slightly above asymptote)
- At $x = 3$: $y = 2^{3-3} + 5 = 2^0 + 5 = 1 + 5 = 6$ (key point)
- At $x = 5$: $y = 2^{5-3} + 5 = 2^2 + 5 = 4 + 5 = 9$
- At $x = 10$: $y = 2^{10-3} + 5 = 2^7 + 5 = 128 + 5 = 133$ (very large)

The curve approaches the horizontal asymptote $y = 5$ from above as $x \to -\infty$, passes through the anchor point $(3, 6)$, and increases without bound as $x \to \infty$.

∴ Shift the graph of $y = 2^x$ right 3 units and up 5 units. The resulting curve has asymptote $y = 5$, key point $(3, 6)$, domain $\mathbb{R}$, and range $(5, \infty)$.

MCQ Practice

Mixed Level — MCQ

1

The fundamental period of $f(x) = 2\cos(x/3)$ is:

Correct!

Incorrect. Correct: $6\pi$

Solution

For any function of the form $f(x) = A\cos(mx)$ or $f(x) = A\sin(mx)$, the period is $P = \frac{2\pi}{|m|}$. The amplitude $A$ does not affect the period; only $m$ does. Here, $m = 1/3$, so $P = \frac{2\pi}{1/3} = 6\pi$.

∴ $6\pi$

2

Which function is odd?

Correct!

Incorrect. Correct: $f(x) = x^3 + \sin x$

Solution

To identify an odd function, check if $f(-x) = -f(x)$. For $f(x) = x^3 + \sin x$: $f(-x) = (-x)^3 + \sin(-x) = -x^3 - \sin x = -(x^3 + \sin x) = -f(x)$. This confirms it is odd.

∴ $f(x) = x^3 + \sin x$

3

The vertex of the parabola $y = x^2 - 6x + 5$ is:

Correct!

Incorrect. Correct: $(3, -4)$

Solution

For $y = ax^2 + bx + c$, the vertex x-coordinate is $x = -b/2a$. Here, $a = 1$, $b = -6$, $c = 5$. So $x = -(-6)/2(1) = 6/2 = 3$. Substituting into the function: $y = 3^2 - 6(3) + 5 = 9 - 18 + 5 = -4$. Vertex = $(3, -4)$.

∴ $(3, -4)$

4

The horizontal asymptote of $y = \frac{3x^2 + 2}{x^2 - 1}$ is:

Correct!

Incorrect. Correct: $y = 3$

Solution

When degrees of numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients. Here, both have degree 2, with leading coefficients 3 and 1 respectively. Thus, $y = 3/1 = 3$.

∴ $y = 3$

5

Which statement correctly describes the transformation from $y = |x|$ to $y = |x+2| - 3$?

Correct!

Incorrect. Correct: Left 2, Down 3

Solution

Remember the counterintuitive rule: in $f(x + h)$, we shift LEFT by $h$ units. In $f(x) + k$, we shift UP by $k$ units. Here, $(x+2)$ shifts left 2, and $-3$ shifts down 3 (negative vertical shift means down).

∴ Left 2, Down 3

6

The function $f(x) = x^3 + 4x$ is:

Correct!

Incorrect. Correct: Odd

Solution

Test: $f(-x) = (-x)^3 + 4(-x) = -x^3 - 4x = -(x^3 + 4x) = -f(x)$. Since $f(-x) = -f(x)$, the function is odd.

∴ Odd

7

The domain of $f(x) = \sqrt{x-2}$ is:

Correct!

Incorrect. Correct: $[2, \infty)$

Solution

For $\sqrt{x-2}$ to be real, the radicand must be non-negative: $x - 2 \geq 0 \Rightarrow x \geq 2$. In interval notation, this is $[2, \infty)$ (bracket includes 2).

∴ $[2, \infty)$

8

If $f(x) = \frac{1}{x-3}$, the vertical asymptote is:

Correct!

Incorrect. Correct: $x = 3$

Solution

Vertical asymptotes occur where the denominator equals zero (and numerator doesn't). Here, $x - 3 = 0 \Rightarrow x = 3$. So the vertical asymptote is $x = 3$.

∴ $x = 3$

9

The period of $\tan(2x)$ is:

Correct!

Incorrect. Correct: $\pi/2$

Solution

For $\tan(mx)$, the period is $P = \frac{\pi}{|m|}$. With $m = 2$, we get $P = \frac{\pi}{2}$. Remember: tangent has period $\pi$ (half that of sine/cosine) because of its discontinuities.

∴ $\pi/2$

10

The range of $f(x) = 2^x$ is:

Correct!

Incorrect. Correct: $(0, \infty)$

Solution

The function $2^x$ is always positive for all real $x$. As $x \to -\infty$, $2^x \to 0$ (but never reaches 0). As $x \to \infty$, $2^x \to \infty$. Therefore, the range is $(0, \infty)$ (all positive reals).

∴ $(0, \infty)$

Short Questions (2-4 Marks Each)

1

Determine if $f(x) = x^4 + 2x^2$ is even or odd, and explain what this means for its graph.

[2 marks]

Solution

To test, we replace $x$ with $-x$:

$$f(-x) = (-x)^4 + 2(-x)^2 = x^4 + 2x^2 = f(x)$$

Since $f(-x) = f(x)$, the function is EVEN.

This means the graph is symmetric about the y-axis. If you fold the paper along the y-axis, the graph on the left perfectly mirrors the graph on the right.

Graph appearance:


        y
        |
      4 +   •           •
        |  / \         / \
      3 + /   \       /   \
        |/     \     /     \
      2 +       •   •       •
        |        \ /
      1 +————————•————————
        |      
Symmetry axis: y-axis (x = 0)
        | mirror line

∴ Even function; symmetric about the y-axis.

2

Find the period of $f(x) = \sin(4x)$.

[2 marks]

Solution

For $\sin(mx)$, the period is $P = \frac{2\pi}{|m|}$.

Here, $m = 4$, so:
$$P = \frac{2\pi}{4} = \frac{\pi}{2}$$

The function repeats every $\pi/2$ units along the x-axis.

Graph showing period:


        y = sin(4x)
        |
      1 +  •     •     •     •
        | / \   / \   / \   / \
      0 +—•——•—•——•—•——•—•——•— x
        0  π/8 π/4 3π/8 π/2 5π/8
        |   \   |  /   
     -1 +     •  •
        |
        |← One period = π/2

∴ Period = $\frac{\pi}{2}$

3

State the domain of $f(x) = \log(x-1)$.

[2 marks]

Solution

For a logarithm to be defined, the argument must be strictly positive:

$$x - 1 > 0$$
$$x > 1$$

In interval notation: Domain = $(1, \infty)$

Graph showing domain:


        y
        |
      2 +     /
        |    /
      1 +   /
        |  /
      0 + • (1,0) ← passes here
        | |
     -1 + | (asymptote x=1)
        |/
        •(2,-∞)
        |
     Domain: (1, ∞)
     Vertical asymptote at x = 1

∴ Domain = $(1, \infty)$

4

Find the x and y intercepts of $y = x^2 - 3x + 2$.

[2 marks]

Solution

Y-intercept (set $x = 0$):
$$y = 0^2 - 3(0) + 2 = 2$$
Y-intercept: $(0, 2)$

X-intercepts (set $y = 0$):
$$x^2 - 3x + 2 = 0$$
$$(x - 1)(x - 2) = 0$$
$$x = 1 \text{ or } x = 2$$
X-intercepts: $(1, 0)$ and $(2, 0)$

Graph showing intercepts:


        y
        |
      2 +  •(0,2)
        | / \
      1 +/   \
        |/     \
      0 +•—————•— x
        0  1   2
        |

∴ Y-intercept: $(0, 2)$; X-intercepts: $(1, 0)$ and $(2, 0)$

5

Identify the vertical and horizontal asymptotes of $f(x) = \frac{2x}{x+1}$.

[3 marks]

Solution

Vertical Asymptote: Set the denominator to zero:
$$x + 1 = 0$$
$$x = -1$$
Vertical asymptote: $x = -1$

Horizontal Asymptote: Compare degrees (both 1). The ratio of leading coefficients:
$$y = \frac{2}{1} = 2$$
Horizontal asymptote: $y = 2$

Graph showing asymptotes:


        y
        | |
      4 + | /
        | |/
      2 +—•—•——— (horizontal asymptote y=2)
        | | \
      0 +—•———•——— x
        |-1   0
        | |    \
     -2 + |     \
        | | (vertical asymptote x=-1)

∴ Vertical: $x = -1$; Horizontal: $y = 2$

6

How does the graph of $y = (x-2)^2 + 3$ relate to the graph of $y = x^2$?

[2 marks]

Solution

The transformation from $y = x^2$ to $y = (x-2)^2 + 3$ involves:

1. Horizontal shift: The $(x-2)$ means shift RIGHT by 2 units.
2. Vertical shift: The $+3$ means shift UP by 3 units.

The parabola moves 2 units right and 3 units up. The vertex moves from $(0, 0)$ to $(2, 3)$.

Visual comparison:


Original y=x²        Transformed y=(x-2)²+3
        y                    y
        |                    |
      4 +                   4 +     •
        |    •              |    • / \
      3 +   / \             |   / \ 3  \
        |  /   \            |  /   •    \
      2 + /     \           | /    2    \
        |/       \          |/          \
      1 +        •          +           •
        |                   |  ↑        ↑
      0 +————•——— x         | shift shift
        |vertex (0,0)       | right up
        |                   | by 2  by 3
        ↓                   ↓
    Vertex (0,0)       Vertex (2,3)

∴ Shift right 2 units and up 3 units. Vertex moves from $(0,0)$ to $(2,3)$.

7

Test the symmetry of $y = \frac{1}{x}$ about the origin.

[2 marks]

Solution

To test symmetry about the origin, replace both $x$ with $-x$ and $y$ with $-y$:

Original: $y = \frac{1}{x}$

Replace: $-y = \frac{1}{-x} = -\frac{1}{x}$

Solving for $y$: $y = \frac{1}{x}$

This matches the original equation, so the curve IS symmetric about the origin. (Alternatively, check if $f(-x) = -f(x)$: $f(-x) = \frac{1}{-x} = -\frac{1}{x} = -f(x)$ ✓)

Graph showing origin symmetry:


        y
        |    /
      2 +   /
        |  /
      1 + •
        |/  \ Origin (0,0)
      0 +————•———— x
        |     \ 
     -1 +      •
        |       \
     -2 +        \
        |
     180° rotation about origin
     gives the same curve

∴ Yes, the curve is symmetric about the origin (odd function).

8

Find the range of $f(x) = x^2 + 4$.

[2 marks]

Solution

Since $x^2 \geq 0$ for all real $x$, we have:
$$f(x) = x^2 + 4 \geq 0 + 4 = 4$$

The minimum value is 4 (achieved when $x = 0$).
There is no maximum value; $f(x)$ grows without bound as $x \to \pm\infty$.

Range = $[4, \infty)$

Graph showing range:


        y
     10 +
        |
>      8 + •
       7 + •   •
Range |  6 +    •
       | 5 +    | •
       | 4 +————•———— minimum point
     3 + |   / (at x=0, y=4)
        | |  /
     2 + |/
        |
      0 + x

[All y-values ≥ 4]

∴ Range = $[4, \infty)$

Long Questions (6-8 Marks Each)

1

Sketch the graph of $y = x^2 + 2x - 3$, showing all important features including vertex, intercepts, and axis of symmetry.

[8 marks]

Solution

We'll use the systematic five-step approach to sketch this parabola.

Step 1: Identify the Function Type

This is a quadratic function $y = x^2 + 2x - 3$ with $a = 1$, $b = 2$, $c = -3$.
Since $a = 1 > 0$, the parabola opens UPWARD.

Step 2: Find the Vertex

Using the vertex formula:
$$x = -\frac{b}{2a} = -\frac{2}{2(1)} = -1$$

Substitute $x = -1$ into the function:
$$y = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$$

Vertex: $(-1, -4)$ (This is the minimum point since the parabola opens upward)

Step 3: Find the Intercepts

Y-intercept (set $x = 0$):
$$y = 0^2 + 2(0) - 3 = -3$$
Y-intercept: $(0, -3)$

X-intercepts (set $y = 0$):
$$x^2 + 2x - 3 = 0$$

We need two numbers that multiply to $-3$ and add to $2$. These are $3$ and $-1$.
$$(x + 3)(x - 1) = 0$$
$$x = -3 \text{ or } x = 1$$

X-intercepts: $(-3, 0)$ and $(1, 0)$

Step 4: Identify the Axis of Symmetry

The axis of symmetry is a vertical line through the vertex:
$$x = -1$$

Step 5: Test a Point for Verification

Let's test $x = 2$ (easy calculation):
$$y = 2^2 + 2(2) - 3 = 4 + 4 - 3 = 5$$
Point: $(2, 5)$

By symmetry, there should be a corresponding point at $x = -4$ (equidistant from the axis $x = -1$):
$$y = (-4)^2 + 2(-4) - 3 = 16 - 8 - 3 = 5$$
Point: $(-4, 5)$ ✓ (This confirms our axis of symmetry)

Step 6: Sketch the Graph

Plot the key points:
- Vertex (minimum): $(-1, -4)$
- X-intercepts: $(-3, 0)$ and $(1, 0)$
- Y-intercept: $(0, -3)$
- Check points: $(2, 5)$ and $(-4, 5)$

Complete Graph:


        y
      5 +   •(-4,5)         •(2,5)
        |  / \             / \
      4 + /   \           /   \
        |/     \         /     \
      3 +       •       •       •
        |        \     /
      2 +         •   •
        |          \ /(0,-3)
      1 +           •
        |           |
      0 +———•———•———•———•———•——— x
       -4  -3  -1   0   1   2
        |           |
     -1 +           •
        |           |
     -2 +           |
        |           |
     -3 +           •(0,-3)
        |           |
     -4 +    ▼ VERTEX(-1,-4)
        |           |
        Axis of symmetry: x = -1 (dashed line)

Properties Summary:
- Domain: $\mathbb{R}$ (all real numbers)
- Range: $[-4, \infty)$ (all y-values from -4 upward)
- Vertex (minimum): $(-1, -4)$
- Axis of symmetry: $x = -1$
- Opens: Upward
- X-intercepts: $-3, 1$
- Y-intercept: $-3$

∴ Upward-opening parabola with vertex (-1,-4), x-intercepts at -3 and 1, y-intercept at -3, axis of symmetry at x = -1.

2

Sketch the graph of $y = \frac{x+1}{x-2}$, clearly showing all asymptotes and key points.

[8 marks]

Solution

This is a rational function. We use the complete analysis framework.

Step 1: Find the Domain

The denominator cannot be zero:
$$x - 2 \neq 0$$
$$x \neq 2$$

Domain: $\mathbb{R} - \{2\}$ or $(-\infty, 2) \cup (2, \infty)$

Step 2: Find the Intercepts

Y-intercept (set $x = 0$):
$$y = \frac{0+1}{0-2} = \frac{1}{-2} = -\frac{1}{2}$$
Y-intercept: $(0, -1/2)$

X-intercept (set $y = 0$):
$$\frac{x+1}{x-2} = 0$$

A fraction equals zero only if numerator = 0 (and denominator ≠ 0):
$$x + 1 = 0$$
$$x = -1$$

Check: denominator at $x = -1$: $-1 - 2 = -3 \neq 0$ ✓
X-intercept: $(-1, 0)$

Step 3: Find the Asymptotes

Vertical Asymptote: Where denominator = 0
$$x - 2 = 0$$
$$x = 2$$ (Vertical asymptote)

Horizontal Asymptote: Compare degrees of numerator (1) and denominator (1). Since degrees are equal, the asymptote is the ratio of leading coefficients:
$$y = \frac{1}{1} = 1$$ (Horizontal asymptote)

Step 4: Test Sign in Each Region

The vertical asymptote at $x = 2$ divides the x-axis into two regions:

Region 1: $x < 2$ (test $x = 0$)
$$y = \frac{0+1}{0-2} = -\frac{1}{2}$$ (negative)

Region 2: $x > 2$ (test $x = 3$)
$$y = \frac{3+1}{3-2} = \frac{4}{1} = 4$$ (positive)

Step 5: Behavior Near Asymptotes

As $x \to 2^-$ (approaching 2 from the left):
Numerator approaches 3, denominator approaches 0 from the negative side
$$y \to -\infty$$

As $x \to 2^+$ (approaching 2 from the right):
Numerator approaches 3, denominator approaches 0 from the positive side
$$y \to +\infty$$

As $x \to \infty$:
$$y = \frac{x+1}{x-2} \approx \frac{x}{x} = 1$$
Approaches $y = 1$ from above

As $x \to -\infty$:
$$y \approx 1$$
Approaches $y = 1$ from below

Step 6: Plot Points and Sketch

Key points to plot:
- X-intercept: $(-1, 0)$
- Y-intercept: $(0, -1/2)$
- Test point: $(3, 4)$
- Test point: $(1, -2)$ [from $y = \frac{1+1}{1-2} = -2$]
- Asymptotes: $x = 2$ (vertical, dashed), $y = 1$ (horizontal, dashed)

Complete Graph:


        y
      5 +                    /
        |                   /
      4 +                 •(3,4)
        |              /
      3 +            /
        |          /
      2 +        /
        |      /
      1 +—————•——— (horizontal asymptote y=1)
        |   /   \
      0 +—•———•———•———— x
        |-1  0   2
     -1 + |  •(1,-2)
        | | /
     -2 + |/
        | |(vertical asymptote x=2)
     -3 +
        |
   LEFT BRANCH         RIGHT BRANCH
   (x < 2):            (x > 2):
   - Approaches        - Emerges from +∞
     y = 1 from below  - Approaches y = 1
   - Curves down       - Curves down
     to -∞ at x=2      from +∞ at x=2

Summary of Properties:
- Domain: All real numbers except $x = 2$
- Range: All real numbers except $y = 1$
- Vertical asymptote: $x = 2$
- Horizontal asymptote: $y = 1$
- X-intercept: $(-1, 0)$
- Y-intercept: $(0, -1/2)$
- No symmetry (neither even nor odd)
- Left branch: increasing as $x \to 2^-$
- Right branch: increasing as $x \to \infty$

∴ Rational function with vertical asymptote at x=2, horizontal asymptote at y=1, x-intercept at (-1,0), y-intercept at (0,-1/2), with two separate branches.

3

Describe the transformation from $y = \sin x$ to $y = 2\sin(x - \pi/3) + 1$, and sketch both graphs on the same axes for $0 \leq x \leq 2\pi$.

[8 marks]

Solution

We analyze how each component of the transformed function modifies the original sine curve.

Step 1: Identify the Transformations

The function $y = 2\sin(x - \pi/3) + 1$ has three transformations applied to $y = \sin x$:

1. Horizontal Shift: $\sin(x - \pi/3)$ means shift RIGHT by $\pi/3$ units
2. Vertical Stretch: Coefficient 2 means multiply all y-values by 2 (stretches vertically)
3. Vertical Shift: $+1$ means shift UP by 1 unit

Step 2: Properties of the Original Function $y = \sin x$

For $0 \leq x \leq 2\pi$:
- Period: $2\pi$
- Amplitude: 1
- Key points: $(0,0)$, $(\pi/2, 1)$, $(\pi, 0)$, $(3\pi/2, -1)$, $(2\pi, 0)$
- Vertical asymptote: none
- Range: $[-1, 1]$

Step 3: Apply Transformations Step-by-Step

Transformation 1: Apply the Horizontal Shift $y = \sin(x - \pi/3)$

Replace all $x$ values with $x - \pi/3$. Key points shift RIGHT by $\pi/3$:
- $(0, 0) \to (\pi/3, 0)$
- $(\pi/2, 1) \to (\pi/2 + \pi/3, 1) = (5\pi/6, 1)$
- $(\pi, 0) \to (\pi + \pi/3, 0) = (4\pi/3, 0)$
- $(3\pi/2, -1) \to (3\pi/2 + \pi/3, -1) = (11\pi/6, -1)$
- $(2\pi, 0) \to (2\pi + \pi/3, 0) = (7\pi/3, 0)$ (outside our interval)

Transformation 2: Apply the Vertical Stretch $y = 2\sin(x - \pi/3)$

Multiply all y-values by 2:
- $(\pi/3, 0) \to (\pi/3, 0)$
- $(5\pi/6, 1) \to (5\pi/6, 2)$
- $(4\pi/3, 0) \to (4\pi/3, 0)$
- $(11\pi/6, -1) \to (11\pi/6, -2)$

Transformation 3: Apply the Vertical Shift $y = 2\sin(x - \pi/3) + 1$

Add 1 to all y-values:
- $(\pi/3, 0) \to (\pi/3, 1)$
- $(5\pi/6, 2) \to (5\pi/6, 3)$
- $(4\pi/3, 0) \to (4\pi/3, 1)$
- $(11\pi/6, -2) \to (11\pi/6, -1)$

Step 4: Properties of the Transformed Function

- Amplitude: 2 (doubled)
- Period: $2\pi$ (unchanged)
- Vertical Shift: +1 (moves up)
- Horizontal Shift: $\pi/3$ to the right
- Range: $[-1, 3]$
- Center Line: $y = 1$ (instead of $y = 0$)

Step 5: Complete Comparison Graph


        y
      3 +        •(5π/6, 3)
        |       / \
      2 +     /     \
        |    /       \
      1 +—•———————•———— CENTER LINE (y=1)
        |/ π/3  4π/3  \
      0 +•—————•———•———•—•— x
        0    π/2  π  3π/2 2π
     -1 +        |       |
        |        |       •(11π/6,-1)
     -2 +

ORIGINAL (y = sin x):
      1 +  •(π/2,1)
        | / \
      0 +•———•———•———— x
        0   π    2π
     -1 +    •(3π/2,-1)

TRANSFORMED (y = 2sin(x-π/3)+1):
      3 +       •(5π/6,3)
        |      / \
      2 +    /     \
        |   /       \
      1 +—•———————•————
        |/  π/3  4π/3 \
      0 +—•———•———•———•— x
        0   π   3π/2  2π
     -1 +            •(11π/6,-1)

KEY TRANSFORMATIONS:
1. Shift RIGHT by π/3
2. Stretch vertically by factor 2
3. Shift UP by 1

Result: Taller curve (amplitude 2), shifted right and up,
oscillating around y=1 instead of y=0

Detailed Comparison Table:

| Feature | Original $y = \sin x$ | Transformed $y = 2\sin(x - \pi/3) + 1$ |
|---------|-----|-----|
| Amplitude | 1 | 2 |
| Period | $2\pi$ | $2\pi$ |
| Center line | $y = 0$ | $y = 1$ |
| Zero at | $x = 0, \pi, 2\pi$ | $x = \pi/3, 4\pi/3$ |
| Maximum | at $x = \pi/2$ (value 1) | at $x = 5\pi/6$ (value 3) |
| Minimum | at $x = 3\pi/2$ (value -1) | at $x = 11\pi/6$ (value -1) |
| Range | $[-1, 1]$ | $[-1, 3]$ |

∴ Shift right $\pi/3$, stretch vertically by factor 2, shift up 1 unit. Transformed curve has amplitude 2, range [-1,3], center line at y=1, with rightmost zero at x=π/3.

One-Page Revision Cheat Sheet

1

Five-Step Sketching Framework: Domain → Intercepts → Symmetry → Asymptotes → Plot Points → Connect. This order is essential and never changes.

2

Even Function ($f(-x) = f(x)$): Symmetric about y-axis. Examples: $x^2, \cos x, |x|$. Mirror the right side to the left.

3

Odd Function ($f(-x) = -f(x)$): Symmetric about origin. Examples: $x^3, \sin x, \frac{1}{x}$. Rotate 180° around origin. Always passes through origin.

4

Period Formulas: $\sin(mx)$ and $\cos(mx)$: $P = \frac{2\pi}{|m|}$. $\tan(mx)$: $P = \frac{\pi}{|m|}$. (Tangent period is half that of sine!)

5

Vertex of Parabola $ax^2+bx+c$: $x = -\frac{b}{2a}$, then substitute to find y. This formula is non-negotiable. Check the negative sign!

6

Horizontal Asymptote (Rational Functions): If degree of numerator < degree of denominator, $y = 0$. If equal degrees, $y = \frac{\text{leading coeff numerator}}{\text{leading coeff denominator}}$. If numerator degree > denominator degree, no horizontal asymptote.

7

Vertical Asymptote (Rational Functions): Set denominator = 0 (check numerator ≠ 0 at that point). This is where the function is undefined and shoots to ±∞.

8

Transformation Paradox - THE SHIFT RULE: In $f(x - h)$, shift RIGHT. In $f(x + h)$, shift LEFT. This opposite direction rule confuses many! The negative makes it shift the opposite direction.

9

Amplitude vs Period: These are different! Amplitude = vertical stretch factor. Period = horizontal repeat distance. In $2\sin(3x)$: amplitude 2, period $2\pi/3$.

10

Domain Restrictions (Three Enemies): (1) Denominators ≠ 0, (2) Even roots ≥ 0, (3) Logarithm arguments > 0. Always check these three.

11

Range Finding Technique: Set $y = f(x)$, solve for $x$ in terms of $y$. Restrictions on $y$ values (where $x$ doesn't exist or is undefined) show excluded range values.

12

Holes vs Asymptotes: A hole is a removable discontinuity (like $(x-2)/(x-2)$ simplifies to 1 with a hole at $x=2$). An asymptote is non-removable (denominator never cancels).

13

Common Function Quick Sketches: Linear ($y=mx+b$): straight line. Quadratic ($ax^2+bx+c$): parabola, opens up if $a>0$, down if $a<0$. Exponential ($a^x$): always positive, passes $(0,1)$, asymptote $y=0$. Logarithm ($\log_a x$): passes $(1,0)$, asymptote $x=0$, domain $(0,\infty)$.

14

Testing for Symmetry (3 Ways): Y-axis symmetry: replace $x$ with $-x$, equation unchanged. X-axis symmetry: replace $y$ with $-y$, equation unchanged (rare for functions). Origin symmetry: replace both $x→-x$ and $y→-y$, equation unchanged.

15

Monotonicity: Increasing = goes up as $x$ increases. Decreasing = goes down as $x$ increases. Use test points or calculus (if learned) to determine on each interval.