U2 Question 1 AI SL Paper 2

Quadratic Function and Graph Features

IB Mathematics SL • Quadratic Functions • Graph Sketching • Symmetry • Tangents • Increasing and Decreasing

Lesson Overview

In this lesson, we analyse the graph of a quadratic function written in factorised form. You will identify the y-intercept, use the vertex to find the axis of symmetry, use symmetry to determine another solution, find the x-intercepts, sketch the graph, and interpret the gradient of a tangent.

The graph of the quadratic function

$f(x)=\frac{1}{2}(x-2)(x+8)$

intersects the y-axis at $(0,c)$ .

⭐ Key Concepts

For a quadratic in factorised form $f(x)=a(x-r_1)(x-r_2)$ :

$\text{x-intercepts are }x=r_1\text{ and }x=r_2$

$\text{axis of symmetry }x=\frac{r_1+r_2}{2}$

$\text{vertex lies on the axis of symmetry}$

The y-intercept is found by substituting $x=0$ .
The graph is symmetric about its axis of symmetry.
If the vertex is a minimum point, the parabola opens upwards.
A horizontal tangent has gradient $0$ .
If $f'(a)<0$ , the function is decreasing at $x=a$ .

📘 Clear IB-Style Explanation

Quadratic questions often require you to connect algebraic features with graphical features. A function written in factorised form gives the x-intercepts immediately, while the vertex and axis of symmetry help you complete the sketch accurately.

In IB questions, always relate the equation to the graph:

substitute $x=0$ to find the y-intercept,
use the vertex to identify the axis of symmetry,
use symmetry to find missing coordinates,
interpret the sign of the derivative to decide whether the graph is increasing or decreasing.

📌 Worked Example 1 — Finding the y-Intercept

The graph of

$f(x)=\frac{1}{2}(x-2)(x+8)$

intersects the y-axis at $(0,c)$ .

Question: Find the value of $c$ .

Solution

The y-intercept occurs when $x=0$ , so

$c=f(0)$

$f(0)=\frac{1}{2}(0-2)(0+8)$

$=\frac{1}{2}(-2)(8)$

$=-8$

Therefore

$c=-8$

📊 Marks: M1 A1

📌 Worked Example 2 — Axis of Symmetry

The vertex of the function is

$(-3,-12.5)$

Question: Write down the equation for the axis of symmetry of the graph.

Solution

The axis of symmetry is the vertical line passing through the vertex.

$x=-3$

📊 Marks: A1

📌 Worked Example 3 — Using Symmetry

The equation

$f(x)=12$

has two solutions. The first solution is

$x=-10$

Question: Use the symmetry of the graph to show that the second solution is $x=4$ .

Solution

The axis of symmetry is

$x=-3$

The point $x=-10$ is

$-3-(-10)=7$

units to the left of the axis of symmetry.

By symmetry, the second solution must be 7 units to the right of the axis.

$-3+7=4$

So the second solution is

$x=4$

📊 Marks: M1 A1

📌 Worked Example 4 — x-Intercepts

Question: Write down the x-intercepts of the graph.

Solution

The x-intercepts occur when

$f(x)=0$

Since

$\frac{1}{2}(x-2)(x+8)=0$

we have

$x-2=0 \quad \text{or} \quad x+8=0$

$x=2 \quad \text{or} \quad x=-8$

Therefore the x-intercepts are

$(-8,0)\quad \text{and}\quad (2,0)$

📊 Marks: A1 A1

📌 Worked Example 5 — Sketching the Graph

Question: On graph paper, draw the graph of $y=f(x)$ for $-10\le x\le 4$ and $-14\le y\le 14$ .

Solution

To sketch the graph accurately, plot the key features:

$\text{vertex }(-3,-12.5)$

$\text{y-intercept }(0,-8)$

$\text{x-intercepts }(-8,0)\text{ and }(2,0)$

Since the coefficient of $x^2$ is positive, the parabola opens upwards.

📊 Marks: A1 for correct shape and vertex placement

📌 Worked Example 6 — Equation of the Tangent

Let $T$ be the tangent at

$x=-3$

(f.i) Write down the equation of $T$ .

Solution

At the vertex, the tangent is horizontal, so its gradient is $0$ .

Since the vertex is

$(-3,-12.5)$

the tangent is the horizontal line

$y=-12.5$

📊 Marks: A1

📌 Worked Example 7 — Drawing the Tangent

(f.ii) Draw the tangent $T$ on your graph.

Solution

Draw the horizontal line

$y=-12.5$

touching the graph at the vertex.

📊 Marks: A1

📌 Worked Example 8 — Increasing or Decreasing

Given

$f(a)=5.5 \quad \text{and} \quad f'(a)=-6$

Question: State whether the function $f$ is increasing or decreasing at $x=a$ . Give a reason for your answer.

Solution

Since

$f'(a)=-6<0$

the gradient of the tangent is negative at $x=a$ .

Therefore, the function is

$\text{decreasing at }x=a$

📊 Marks: A1 A1

⚠ Common Mistakes

Forgetting to substitute $x=0$ when finding the y-intercept.
Confusing the axis of symmetry with the y-axis.
Listing x-intercepts as x-values only when coordinates are required.
Forgetting that the tangent at the vertex of a parabola can be horizontal.
Saying a function is increasing when the derivative is negative.

📘 IB Exam Tips

Use the equation form to identify graph features efficiently.
Always mark the vertex clearly before sketching a parabola.
Use symmetry to save time when finding missing points.
When interpreting $f'(x)$ , focus on the sign of the derivative.
Give graph answers in coordinates where appropriate.

🧪 Challenge Problem

The graph of the quadratic function

$g(x)=\frac{1}{2}(x-4)(x+6)$

intersects the y-axis at $(0,d)$ .

(a) Find the value of $d$ .

(b) Write down the x-intercepts of the graph.

(c) Find the equation of the axis of symmetry.

(d) Find the coordinates of the vertex.

✅ Self-Practice

Consider the quadratic function

$h(x)=(x-1)(x+5)$

(a) Find the y-intercept.

(b) Write down the x-intercepts.

(c) Find the axis of symmetry.

(d) Find the coordinates of the vertex.

(e) State whether the function is increasing or decreasing at a point where $h'(x)=-4$ .

Show Solutions

Challenge Problem

(a)

$d=g(0)$

$g(0)=\frac{1}{2}(0-4)(0+6)$

$=\frac{1}{2}(-4)(6)=-12$

(b)

$g(x)=0$

$\frac{1}{2}(x-4)(x+6)=0$

$x=4 \quad \text{or} \quad x=-6$

So the x-intercepts are

$(-6,0)\quad \text{and}\quad (4,0)$

(c)

$x=\frac{-6+4}{2}=-1$

So the axis of symmetry is

$x=-1$

(d)

Substitute $x=-1$ into $g(x)$ :

$g(-1)=\frac{1}{2}(-1-4)(-1+6)$

$=\frac{1}{2}(-5)(5)$

$=-12.5$

So the vertex is

$(-1,-12.5)$

Self-Practice

(a)

$h(0)=(0-1)(0+5)$

$=(-1)(5)=-5$

So the y-intercept is

$(0,-5)$

(b)

$h(x)=0$

$(x-1)(x+5)=0$

$x=1 \quad \text{or} \quad x=-5$

So the x-intercepts are

$(-5,0)\quad \text{and}\quad (1,0)$

(c)

$x=\frac{-5+1}{2}=-2$

So the axis of symmetry is

$x=-2$

(d)

$h(-2)=(-2-1)(-2+5)$

$=(-3)(3)=-9$

So the vertex is

$(-2,-9)$

(e)

Since

$h'(x)=-4<0$

the gradient is negative, so the function is

$\text{decreasing}$

Quadratic Function and Graph Features

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example 1 — Finding the y-Intercept

Solution

📌 Worked Example 2 — Axis of Symmetry

Solution

📌 Worked Example 3 — Using Symmetry

Solution

📌 Worked Example 4 — x-Intercepts

Solution

📌 Worked Example 5 — Sketching the Graph

Solution

📌 Worked Example 6 — Equation of the Tangent

Solution

📌 Worked Example 7 — Drawing the Tangent

Solution

📌 Worked Example 8 — Increasing or Decreasing

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

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U2 Question 1 AI SL Paper 2

Quadratic Function and Graph Features

Lesson Overview

⭐ Key Concepts

📘 Clear IB-Style Explanation

📌 Worked Example 1 — Finding the y-Intercept

Solution

📌 Worked Example 2 — Axis of Symmetry

Solution

📌 Worked Example 3 — Using Symmetry

Solution

📌 Worked Example 4 — x-Intercepts

Solution

📌 Worked Example 5 — Sketching the Graph

Solution

📌 Worked Example 6 — Equation of the Tangent

Solution

📌 Worked Example 7 — Drawing the Tangent

Solution

📌 Worked Example 8 — Increasing or Decreasing

Solution

⚠ Common Mistakes

📘 IB Exam Tips

🧪 Challenge Problem

✅ Self-Practice

Related

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