A. Quadratic Functions · Quadratic Functions

Sketching from Factored Form: Using the Zeros to Find the Vertex

A2.9About 65 minutes

Learning goals

I can read the zeros and direction of opening from f(x) = a(x − r)(x − s).
I can find the axis of symmetry and the vertex from the zeros.
I can sketch a quadratic function given in factored form.
I can factor standard form first when needed, and convert to vertex form.

Words to know

factored form
f(x) = a(x − r)(x − s), where r and s are the zeros.

zero
An input that gives an output of 0; an x-intercept of the graph.

axis of symmetry
For factored form, the line x = (r + s) ÷ 2, halfway between the zeros.

double zero
A repeated zero, as in a(x − r)²; the vertex touches the x-axis at r.

The zeros tell you a lot

A soccer ball is kicked from the ground and lands 40 m away. Before you know anything else, you know two points on its path: where it started and where it landed. Because a parabola is symmetric, the highest point must be exactly halfway between them.

That is the big idea of the factored form of a quadratic function:

f(x) = a(x − r)(x − s)

It shows the zeros r and s directly. From them you can find the axis of symmetry and the vertex, and sketch the graph.

Remember from Grade 10 and the sibling unit

A zero of a function is an x-value where f(x) = 0: an x-intercept of the graph. If a product is 0, one of its factors is 0. So a(x − r)(x − s) = 0 when x = r or x = s. Factoring standard form is taught in Quadratic Expressions and Equations.

Reading factored form

For f(x) = a(x − r)(x − s):

Zeros: x = r and x = s. Watch the signs: (x + 3) = (x − (−3)) gives the zero −3.
Axis of symmetry: halfway between the zeros, x = (r + s) ÷ 2.
Vertex: substitute the axis value into f to get the y-coordinate.
Direction: up if a > 0, down if a < 0.
y-intercept: f(0) = a(0 − r)(0 − s) = a·r·s.

If r = s, the function is f(x) = a(x − r)². It has one zero, and the vertex (r, 0) sits on the x-axis.

For f(x) = −2(x + 3)(x − 1), the zeros are −3 and 1. The axis is halfway, at x = −1. Substituting gives the vertex (−1, 8). The y-intercept is −2(3)(−1) = 6.

Sketching from factored form

Plot the zeros r and s.
Draw the axis x = (r + s) ÷ 2.
Find the vertex by substituting the axis value.
Find the y-intercept a·r·s, and its mirror point across the axis.
Sketch a smooth parabola through these points.

Sketching f(x) = (x − 1)(x − 5) in four steps: zeros 1 and 5, axis x = 3, vertex (3, −4), then the y-intercept (0, 5) and its mirror point (6, 5) before drawing the curve.

Worked example

Example 1: f(x) = (x − 1)(x − 5)

Sketch f(x) = (x − 1)(x − 5). State the zeros, axis, vertex and y-intercept.

Zeros: 1 and 5.
Axis: x = (1 + 5) ÷ 2 = 3.
Vertex: f(3) = (3 − 1)(3 − 5) = (2)(−2) = −4, so (3, −4). a = 1 > 0, so it opens up and −4 is the minimum.
y-intercept: f(0) = (−1)(−5) = 5. Mirror point across x = 3: (6, 5).
Sketch through (1, 0), (5, 0), (3, −4), (0, 5) and (6, 5).

Zeros 1 and 5; axis x = 3; vertex (3, −4); y-intercept 5.

Worked example

Example 2: f(x) = −2(x + 3)(x − 1)

Find the key features of f(x) = −2(x + 3)(x − 1) and state the range.

Zeros: x + 3 = 0 gives −3; x − 1 = 0 gives 1.
Axis: x = (−3 + 1) ÷ 2 = −1.
Vertex: f(−1) = −2(2)(−2) = 8, so (−1, 8). a = −2 < 0: opens down, so 8 is the maximum.
y-intercept: f(0) = −2(3)(−1) = 6.
Range: {y ∈ ℝ | y ≤ 8}.

Zeros −3 and 1; axis x = −1; vertex (−1, 8); y-intercept 6; R = {y ∈ ℝ | y ≤ 8}.

Worked example

Example 3: Factor first, then sketch

Sketch f(x) = x² − 2x − 8 by first writing it in factored form. Then write it in vertex form.

Find two numbers with product −8 and sum −2: −4 and 2. So f(x) = (x − 4)(x + 2).
Zeros 4 and −2; axis x = (4 + (−2)) ÷ 2 = 1.
Vertex: f(1) = (−3)(3) = −9, so (1, −9). y-intercept: −8 (from c).
Vertex form: a = 1, vertex (1, −9), so f(x) = (x − 1)² − 9. Check: x² − 2x + 1 − 9 = x² − 2x − 8 ✓.

f(x) = (x − 4)(x + 2) = (x − 1)² − 9.

Worked example

Example 4: A context

A frog jumps. Its height in centimetres is h(x) = −0.05x(x − 40), where x is the horizontal distance in centimetres from where it took off. How far does it jump, how high does it get, and what are the domain and range in this context?

Zeros: x = 0 (take-off) and x = 40 (landing). The jump is 40 cm long.
Axis: x = (0 + 40) ÷ 2 = 20.
h(20) = −0.05(20)(20 − 40) = −0.05(20)(−20) = 20. The highest point is 20 cm up, halfway along the jump.
Reasonable? A 40 cm long, 20 cm high hop is believable for a small frog. ✓
D = {x ∈ ℝ | 0 ≤ x ≤ 40}; R = {h ∈ ℝ | 0 ≤ h ≤ 20}.

The jump is 40 cm long and 20 cm high. D = {x ∈ ℝ | 0 ≤ x ≤ 40}; R = {h ∈ ℝ | 0 ≤ h ≤ 20}.

Just for fun

A parabola in factored form is like a well-planned road trip: you know where you start, you know where you finish, and the highlight is exactly halfway.

Hands-on activity

Fold to find the axis

You need: graph paper, a pencil, and a ruler.

Draw axes on graph paper. Choose two zeros, for example r = −2 and s = 6. Mark them on the x-axis.
Fold the paper so that the two zero marks land exactly on top of each other. Crease it and unfold. The crease is a vertical line. Where does it cross the x-axis? That is the axis of symmetry.
Choose a = −0.5. Your function is f(x) = −0.5(x + 2)(x − 6). Compute the vertex and y-intercept.
Plot the zeros, vertex, y-intercept and its mirror point. Sketch the parabola. Fold along the crease again: the two halves should match.
Repeat with your own choice of r, s and a.

What to look for: the crease is at x = (−2 + 6) ÷ 2 = 2. f(2) = −0.5(4)(−4) = 8, so the vertex is (2, 8). The y-intercept is −0.5(2)(−6) = 6, with mirror point (4, 6).

Your turn

For f(x) = (x − 2)(x − 6), state the zeros, axis, vertex and y-intercept.

For f(x) = −(x + 1)(x − 5), state the zeros, vertex and y-intercept.

For f(x) = 2(x + 4)(x + 2), state the zeros and the vertex.

For f(x) = 0.5x(x − 8), state the zeros and the minimum value.

Factor f(x) = x² − 6x + 5, then find the vertex.

Describe the graph of f(x) = −3(x − 2)².

Write f(x) = (x − 1)(x − 5) in vertex form.

More practice: warm-up

State the zeros of f(x) = (x − 3)(x + 7).

What is the axis of symmetry if the zeros are −2 and 6?

Does f(x) = −4(x − 1)(x − 2) open up or down?

State the y-intercept of f(x) = (x − 2)(x − 3).

State the zeros of f(x) = x(x + 5).

More practice: on your own

For f(x) = −0.5(x − 3)(x + 5), find the zeros, vertex, y-intercept and range.

Write f(x) = 2x² + 4x − 6 in factored form and find the vertex.

Find the vertex of f(x) = x² − 9 using factored form.

A parabola has zeros −2 and 4 and passes through (0, −16). Find its equation in factored form and its vertex.

The arch of a footbridge is modelled by h(x) = −0.04x(x − 30), in metres. How wide is the arch at its base and how high is it?

Challenge

Expand f(x) = a(x − r)(x − s) and use h = −b ÷ (2a) to prove that the axis of symmetry is x = (r + s) ÷ 2.

Compare f(x) = (x − 1)(x − 7) and g(x) = −(x − 1)(x − 7). What do they share, and how do they differ?

f(x) = a(x − 2)(x − 8) has a maximum value of 18. Find a and write f in vertex form.

Check your understanding

Quiz

What are the zeros of f(x) = 3(x + 2)(x − 5)?

What is the axis of symmetry of f(x) = (x − 1)(x + 7)?

What is the vertex of f(x) = (x − 2)(x − 4)?

What is the y-intercept of f(x) = −2(x − 1)(x + 3)?

How many zeros does f(x) = (x − 4)² have?

What is the maximum value of f(x) = −x(x − 10)?

For parents and teachers

Big idea: expectation A2.9 asks students to sketch graphs of quadratic functions in factored form f(x) = a(x − r)(x − s) by using the x-intercepts to determine the vertex.

How to help: ask "Where does it cross the x-axis? Where is halfway? How high (or low) is it there?" Those three questions give a complete sketch. Encourage a reasonableness check in context problems, as in Example 4.

Common difficulties:

Sign errors reading zeros: (x + 3) gives −3.
Forgetting to substitute the axis value to get the vertex's y-coordinate (giving only the x).
Forgetting a in the y-intercept (a·r·s).
Assuming every quadratic can be written in factored form: some have no real zeros (see Comparing the Three Forms).