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Math Coach Amy
Amy Ferguson Moncure

IB AA · 3B Sketching Graphs

Zeros and Roots

A zero is an x-value that makes the output 0.

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Key Ideas Easy Medium Hard

A zero is where the graph meets the x-axis

What to Remember

Factor

If the polynomial factors, set each factor equal to 0.

A square root in the equation

Let u = square root of x. Solve for u, then square to find x. Check that x is not negative.

Absolute value

Split into the inside-nonnegative case and the inside-negative case. Keep only the answers that match the case.

No factors

Draw the graph and read each x-intercept to 3 significant figures.

Easy

Easy

Factor a quadratic.

Example
Solve x² − 5x + 6 = 0.

x² − 5x + 6 = 0

(x − 2)(x − 3) = 0

x = 2 or x = 3

Answer: x = 2 or x = 3

1
Solve x² − x − 12 = 0.
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x² − x − 12 = 0

(x − 4)(x + 3) = 0

x = 4 or x = −3

Answer: x = 4 or x = −3

2
Solve x² − 9 = 0.
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x² − 9 = 0

(x − 3)(x + 3) = 0

x = 3 or x = −3

Answer: x = 3 or x = −3

3
Solve x² + 4x = 0.
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x² + 4x = 0

x(x + 4) = 0

x = 0 or x = −4

Answer: x = 0 or x = −4

4
Solve x² − 7x + 10 = 0.
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(x − 2)(x − 5) = 0

x = 2 or x = 5

Answer: x = 2 or x = 5

5
Solve x² + x − 6 = 0.
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(x + 3)(x − 2) = 0

x = −3 or x = 2

Answer: x = −3 or x = 2

Medium

Medium

A cubic, or a substitution.

Example
Find the zeros of f(x) = x³ − 6x² + 11x − 6.

(x − 1)(x − 2)(x − 3) = 0

x = 1, x = 2, or x = 3

Answer: x = 1, 2, or 3

1
Find the zeros of f(x) = x³ − x.
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x(x − 1)(x + 1) = 0

x = −1, 0, or 1

Answer: x = −1, 0, or 1

2
Solve x − 5square root of x + 6 = 0.
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Let u = square root of x. Then u² − 5u + 6 = 0.

(u − 2)(u − 3) = 0

u = 2 or u = 3

x = 4 or x = 9

Answer: x = 4 or x = 9

3
Solve x + 4 = 5square root of x.
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Let u = square root of x.

u² − 5u + 4 = 0

(u − 1)(u − 4) = 0

x = 1 or x = 16

Answer: x = 1 or x = 16

4
Solve |9 − x²| = x².
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Case 1: x² ≤ 9. Then 9 − x² = x², so 9 = 2x² and x² = 92.

x = 3square root of 22 or x = −3square root of 22

Both squares are less than 9, so both answers count.

Case 2: x² > 9. Then x² − 9 = x², which says −9 = 0. There is no solution in this case.

Answer: x = −3square root of 22 or x = 3square root of 22

5
Solve x³ − 4x = 0.
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x(x² − 4) = 0

x(x − 2)(x + 2) = 0

x = −2, 0, or 2

Answer: x = −2, 0, or 2

Hard

Hard

An absolute value, or a root to 3 significant figures.

Example
Find the real root of x³ + x − 4 = 0 to 3 significant figures.

The graph of y = x³ + x − 4 crosses the x-axis once.

x = 1.38

Answer: 1.38

1
Find the root of eˣ + x − 4 = 0 to 3 significant figures.
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Graph y = eˣ + x − 4.

x = 1.07

Answer: 1.07

2
Find the positive root of x² = eˣ − 3 to 3 significant figures, if you rearrange it as x² − eˣ + 3 = 0.
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Graph y = x² − eˣ + 3 and read the positive intercept.

x = 1.87

Answer: 1.87

3
Solve |4 − x²| = 5.
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If x² ≤ 4, 4 − x² = 5 gives x² = −1, which is impossible for real x.

If x² ≥ 4, x² − 4 = 5, so x² = 9.

x = 3 or x = −3, and both have x² ≥ 4.

Answer: x = −3 or x = 3

4
Solve |x² − 16| = 9.
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Case 1: x² ≥ 16. Then x² − 16 = 9, so x² = 25.

x = 5 or x = −5, and both satisfy x² ≥ 16.

Case 2: x² < 16. Then 16 − x² = 9, so x² = 7.

x = square root of 7 or x = −square root of 7, and both satisfy x² < 16.

Answer: x = −5, x = −square root of 7, x = square root of 7, or x = 5

5
Find the real root of x³ − x − 1 = 0 to 3 significant figures.
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The graph of y = x³ − x − 1 crosses the x-axis once.

x = 1.32

Answer: 1.32

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