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Amy Ferguson Moncure

Algebra 1 · Unit 8

Standard Form and Factored Form

Connect standard form and factored form of a quadratic to intercepts, the vertex, and the graph.

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Key Ideas Easy Medium Hard Super Challenge
What to Remember

Standard form

f(x) = ax² + bx + c. Easy y-intercept: (0, c). Axis x = −b2a.

Factored form

f(x) = a(x − r)(x − s). Easy zeros: (r, 0) and (s, 0). Axis halfway between the zeros.

Same graph

Expanding factored form, or factoring standard form, does not change the parabola. Check by matching a, zeros, and f(0).

Sketch from either

Plot the intercepts you can see, then the vertex on the axis of symmetry.

Level 1

Easy

Read zeros from factored form, or c from standard form.

Example
f(x) = (x − 2)(x − 6). Name the zeros and the y-intercept.

Zeros x = 2 and x = 6

f(0) = (−2)(−6) = 12

y-intercept (0, 12)

Answer: zeros 2, 6; (0, 12)

1
f(x) = x² − 5x + 4. Expand or factor to find the zeros.
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(x − 1)(x − 4) = 0

Zeros 1 and 4

Answer: 1 and 4

2
f(x) = 2(x + 1)(x − 3). Name the zeros and f(0).
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Zeros −1 and 3

f(0) = 2(1)(−3) = −6

Answer: −1, 3; f(0) = −6

3
Which form shows the y-intercept of f(x) = x² + 3x − 10 at a glance?
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Standard form: c = −10, so (0, −10).

Answer: standard form; (0, −10)

4
Which form shows the zeros of f(x) = (x + 4)(x − 1) at a glance?
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Factored form: x = −4 and x = 1.

Answer: factored form; −4 and 1

5
Expand f(x) = (x − 3)(x + 2).
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x² − x − 6

Answer: x² − x − 6

Level 2

Medium

Convert, then sketch from intercepts and the vertex.

Example
Write f(x) = x² − 8x + 12 in factored form, then name the vertex.

(x − 2)(x − 6)

Zeros 2 and 6, so axis x = 4

f(4) = 16 − 32 + 12 = −4

Vertex (4, −4)

Answer: (x − 2)(x − 6); (4, −4)

1
Write f(x) = (x + 1)(x − 5) in standard form and find the axis.
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x² − 4x − 5

Zeros −1 and 5, axis x = 2

Answer: x² − 4x − 5; x = 2

2
Sketch f(x) = (x − 1)(x − 5) using intercepts and the vertex. Graph only in the answer.
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Zeros (1, 0), (5, 0). f(0) = 5. Vertex x = 3, f(3) = (2)(−2) = −4

Answer: vertex (3, −4)

3
f(x) = 2x² − 8x + 6. Factor completely and name the zeros.
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2(x² − 4x + 3) = 2(x − 1)(x − 3)

Zeros 1 and 3

Answer: 2(x − 1)(x − 3)

4
A graph crosses the x-axis at −2 and 4 and the y-axis at −8. Write a factored form.
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f(x) = a(x + 2)(x − 4)

−8 = a(2)(−4)

−8 = −8a

a = 1

f(x) = (x + 2)(x − 4)

Answer: (x + 2)(x − 4)

5
Match: standard x² + 2x − 8, factored (x + 4)(x − 2). Same parabola?
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(x + 4)(x − 2) = x² + 2x − 8. Yes.

Answer: yes

Level 3

Hard

a ≠ 1, or a missing factor that still matches the graph.

Example
A parabola has zeros 1 and 4 and passes through (2, −6). Write factored and standard form.

f(x) = a(x − 1)(x − 4)

−6 = a(1)(−2)

−6 = −2a

a = 3

f(x) = 3(x − 1)(x − 4)

f(x) = 3(x² − 5x + 4) = 3x² − 15x + 12

Answer: 3(x − 1)(x − 4) = 3x² − 15x + 12

1
f(x) = −1(x + 3)(x − 1). Expand, then find the vertex from the zeros.
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−1(x² + 2x − 3) = −1x² − 2x + 3

Zeros −3 and 1, axis x = −1

f(−1) = −1(2)(−2) = 4

Vertex (−1, 4)

Answer: −x² − 2x + 3; (−1, 4)

2
Why can two different-looking factored forms, 2(x − 3)(x + 1) and (2x − 6)(x + 1), be the same graph?
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(2x − 6)(x + 1) = 2(x − 3)(x + 1)

The 2 is inside one binomial instead of out front. Same product.

Answer: same after distributing

3
f(x) = x² + 6x + 9. Write factored form and explain the graph’s one x-intercept.
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(x + 3)²

A double zero at x = −3. The vertex sits on the x-axis.

Answer: (x + 3)²; one intercept

4
Standard form 2x² − 2x − 12 and factored 2(x − 3)(x + 2). Show they match two ways: expand, and compare f(0).
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2(x² − x − 6) = 2x² − 2x − 12

f(0) = −12 from standard, and 2(−3)(2) = −12 from factored.

Answer: match

5
A graph opens down, zeros at 0 and 6. Write one possible factored form and its standard form.
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Example: f(x) = −1x(x − 6)

f(x) = −1x² + 6x

Answer: −x(x − 6) = −x² + 6x

Super Challenge

Super Challenge

1
A student expands (x − 4)(x − 1) as x² − 4 and graphs zeros at 2 only. Correct both the algebra and the graph plan.
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(x − 4)(x − 1) = x² − 5x + 4, not x² − 4.

Zeros 4 and 1. Axis x = 52.

Vertex x = 52, f = (52 − 4)(52 − 1) = (−32)(32) = −94.

Answer: x² − 5x + 4; zeros 1 and 4

2
Show that factored form makes the axis obvious, while standard form makes f(0) obvious, using f(x) = 2(x − 1)(x − 7).
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Zeros 1 and 7, axis x = 4, no extra work.

Expand: 2(x² − 8x + 7) = 2x² − 16x + 14, so f(0) = 14 is the constant.

Answer: axis x = 4; f(0) = 14

3
Write two different factored forms that expand to the same standard form x² − 9.
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(x − 3)(x + 3)

Also −1(3 − x)(x + 3), which is the same product.

The usual form is (x − 3)(x + 3).

Answer: (x − 3)(x + 3)

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