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Quadratic Relations of the Form y = ax² + bx + c · Quadratic Expressions and Equations

Factored Form: Connecting Factors and Zeros

3.3 3.6 4.1About 70 minutes
Learning goals
  • I can explain why the factors of a quadratic expression give the zeros of its graph.
  • I can find the zeros, axis of symmetry, vertex and y-intercept from y = a(x − r)(x − s).
  • I can factor a relation in standard form and sketch it using intercepts and symmetry.
  • I can write the equation of a parabola from its zeros and one other point.

Words to know

factored form
A quadratic relation written as y = a(x − r)(x − s). Its zeros are r and s.
zeros
The x-values that make y = 0. They are the x-intercepts of the graph.
x-intercept
A point where a graph crosses or touches the x-axis.
axis of symmetry
The vertical line through the vertex that divides a parabola into two mirror-image halves. In factored form it is x = (r + s) ÷ 2.
vertex
The turning point of a parabola: its lowest point (minimum) or highest point (maximum).
standard form
A quadratic relation written as y = ax² + bx + c.

A graph that tells you its own x-intercepts

Open a graphing tool (such as Desmos) or use graph paper and graph y = (x − 1)(x − 5). Where does the parabola cross the x-axis?

It crosses at x = 1 and x = 5, exactly the numbers that appear in the brackets (with the signs flipped). That is no accident. In this lesson you will find out why, and you will use it to sketch parabolas and write their equations.

Remember from Quadratic Relations and Their Graphs

The zeros (x-intercepts) of a parabola are where y = 0. The axis of symmetry is the vertical line through the vertex. The vertex is the minimum point if the parabola opens up (a > 0) and the maximum point if it opens down (a < 0). The y-intercept is where x = 0.

Why the factors give the zeros

On the x-axis, y = 0. For y = (x − 1)(x − 5), that means

(x − 1)(x − 5) = 0.

When a product of two numbers is zero, at least one of them must be zero. So x − 1 = 0 or x − 5 = 0, which gives x = 1 or x = 5.

In general, a quadratic relation in factored form

y = a(x − r)(x − s)

has zeros at x = r and x = s. The value of a tells you the direction of opening and the stretch, just as in vertex form.

FeatureHow to find it from y = a(x − r)(x − s)
Zeros (x-intercepts)x = r and x = s (flip the signs in the brackets)
Axis of symmetryx = (r + s) ÷ 2, halfway between the zeros
Vertexx-coordinate from the axis; substitute it to get the y-coordinate
Max or mina > 0: opens up, minimum. a < 0: opens down, maximum
y-interceptsubstitute x = 0
Everything you need for a sketch comes out of the factored form.
Worked example

Example 1: Sketch from factored form

For y = (x − 1)(x − 5), find the zeros, axis of symmetry, vertex and y-intercept, and sketch the graph.

  1. Zeros: x − 1 = 0 or x − 5 = 0, so x = 1 and x = 5.
  2. Axis of symmetry: halfway between, x = (1 + 5) ÷ 2 = 3.
  3. Vertex: y = (3 − 1)(3 − 5) = (2)(−2) = −4. Vertex (3, −4).
  4. a = 1 > 0, so the parabola opens up and −4 is the minimum value.
  5. y-intercept: y = (0 − 1)(0 − 5) = 5.
  6. Plot the five facts and draw a smooth U through them. By symmetry, (6, 5) is also on the graph.

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

−2−1012345678−6−4−20246810xy(1, 0)(5, 0)vertex (3, −4)(0, 5)x = 3
y = (x − 1)(x − 5). The zeros are 1 and 5, and the axis of symmetry x = 3 is exactly halfway between them.
Worked example

Example 2: A parabola that opens down

Sketch y = −2(x + 1)(x − 3).

  1. Zeros: x + 1 = 0 gives x = −1; x − 3 = 0 gives x = 3. (Careful: (x + 1) means r = −1.)
  2. Axis: x = (−1 + 3) ÷ 2 = 1.
  3. Vertex: y = −2(1 + 1)(1 − 3) = −2(2)(−2) = 8. Vertex (1, 8).
  4. a = −2 < 0, so it opens down and 8 is the maximum value.
  5. y-intercept: y = −2(1)(−3) = 6.

Zeros −1 and 3; axis x = 1; maximum point (1, 8); y-intercept 6.

−3−2−1012345−8−6−4−20246810xy(−1, 0)(3, 0)vertex (1, 8)(0, 6)x = 1
y = −2(x + 1)(x − 3). The negative a flips the parabola so it opens down; the factors still give the zeros −1 and 3.
Worked example

Example 3: Start in standard form

Sketch y = x² − 2x − 8 using intercepts and symmetry.

  1. Factor: two numbers with product −8 and sum −2 are −4 and 2. So y = (x − 4)(x + 2).
  2. Zeros: x = 4 and x = −2.
  3. Axis: x = (4 + (−2)) ÷ 2 = 1.
  4. Vertex: y = (1 − 4)(1 + 2) = (−3)(3) = −9. Vertex (1, −9), a minimum.
  5. y-intercept: from the standard form, c = −8.

Zeros −2 and 4; axis x = 1; vertex (1, −9); y-intercept −8. Factoring turned a standard-form equation into a quick sketch.

Worked example

Example 4: Find the equation from the graph

A parabola has zeros at x = −3 and x = 1 and a y-intercept of −6. Find its equation in factored form and in standard form.

  1. Use the zeros: y = a(x + 3)(x − 1).
  2. Use the y-intercept point (0, −6): −6 = a(0 + 3)(0 − 1) = −3a.
  3. Solve: a = 2.
  4. Factored form: y = 2(x + 3)(x − 1).
  5. Standard form: 2(x² + 2x − 3) = 2x² + 4x − 6.

y = 2(x + 3)(x − 1), or y = 2x² + 4x − 6.

Special cases

  • One zero: y = (x − 2)² = (x − 2)(x − 2) has r = s = 2. The vertex (2, 0) sits right on the x-axis; the parabola touches the axis but does not cross it.
  • A zero at the origin: y = 3x(x − 4) has zeros 0 and 4, because the factor x is the same as (x − 0).
  • No zeros: some parabolas never reach the x-axis, such as y = x² + 4. Their equations do not factor. You will learn more in the lesson on real and non-real roots.
Hands-on activity

Fold to find the axis

You need: graph paper (or plain paper with a grid you draw), a pencil, and Desmos or another free graphing tool if a grown-up says it is OK to use it.

  1. On graph paper, make a table of values for y = (x + 2)(x − 4) using x = −3 to 5, and plot the points. Join them with a smooth curve.
  2. Mark the zeros on the x-axis.
  3. Fold the paper so the two zeros land on top of each other. Unfold. The crease is the axis of symmetry.
  4. Where does the crease cross the curve? That is the vertex. Check by calculating it.
  5. If you have a graphing tool, type y = a(x + 2)(x − 4) and add a slider for a. What changes and what stays the same as a changes?

Model answer: the zeros are −2 and 4, the crease is x = 1, and the vertex is (1, −9) because (1 + 2)(1 − 4) = −9. Changing a stretches or flips the parabola, but the zeros and the axis stay the same.

Just for fun

Why did the parabola and the x-axis stay just friends?

They had only two points in common. (With (x − 2)², it is even worse: they just touch once and move on.)

Your turn

1

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

2

For y = 2(x + 4)(x − 2), find the zeros, axis of symmetry and vertex. Does it open up or down?

3

For y = −(x − 1)(x − 7), find the zeros, vertex and y-intercept.

4

Factor y = x² + 6x + 5, then find the zeros, vertex and y-intercept.

5

Write the equation of a parabola with zeros 2 and 6 and a = 3.

6

A parabola has zeros −2 and 4 and passes through (1, 18). Find its equation.

7

How many zeros does y = (x − 5)² have? Where is its vertex?

More practice: warm-up

1

State the zeros of y = (x − 2)(x − 9).

2

State the axis of symmetry of y = (x − 2)(x − 9).

3

State the zeros of y = 4x(x + 3).

4

Does y = −5(x − 1)(x + 1) open up or down?

5

Find the y-intercept of y = (x − 2)(x − 9).

More practice: on your own

1

Factor y = 2x² − 8x − 10 fully and give the zeros and vertex.

2

Find the zeros and vertex of y = x² − 9.

3

A parabola has zeros −5 and 1 and a maximum value of 18. Find its equation.

4

Find the zeros and vertex of y = −x² + 2x + 15.

5

The height of a ball is h = −5(t + 1)(t − 4), where h is in metres and t ≥ 0 is in seconds. When does the ball hit the ground, and how high was it at t = 0?

Challenge

1

Compare y = (x + 1)(x − 5) and y = −3(x + 1)(x − 5). What is the same and what is different?

2

Factor y = 2x² + x − 6 and find the zeros and the vertex.

3

Find the equation of the parabola with zeros ±3 that passes through (1, 4). Give it in factored and standard form.

Check your understanding

Quiz

1.

What are the zeros of y = (x + 2)(x − 6)?

2.

What is the axis of symmetry of y = 3(x − 1)(x − 7)?

3.

What is the vertex of y = (x − 1)(x − 5)?

4.

Which relation has zeros 4 and −1?

5.

The parabola y = −(x − 2)(x − 8) has:

6.

A parabola has zeros 0 and 6 and passes through (3, −9). Its equation is:

For parents and teachers

Big idea: Expectation 3.3 asks students to discover the connection between the factors of a quadratic expression and the zeros (x-intercepts) of its graph, written as y = a(x − r)(x − s). Together with expectation 4.1, students also find the maximum or minimum from the equation: it lies on the axis of symmetry, halfway between the zeros.

How to help: If you can, look at Desmos (desmos.com, free) together and type y = a(x − 1)(x − 5) with a slider for a. Ask, "What stays the same when a changes?" The folding activity is a quick, physical way to see the axis of symmetry.

Common difficulties:

  • Sign errors: (x + 3) gives the zero x = −3, not 3.
  • Forgetting that the vertex's x-coordinate is the average of the zeros, then substituting it to find y.
  • Not using a given point to find a when writing an equation.