Standard Form Graphs | Math Coach Amy

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Key Ideas Axis of Symmetry Vertex Y-intercept Minimum and Maximum Domain and Range Table of 5 Points Medium Hard

x = −b2a

axis answer: x = …

Axis formula

The axis of symmetry comes from the quadratic formula.

Quadratic formula

x = −b ± square root of b² − 4ac2a

Split off the ± square root. The first part is the axis.

x =−b2a±square root of b² − 4ac2a

Highlighted: x = −b2a

Axis formula

x = −b2a

This is the highlighted part of the quadratic formula.

Vertex

The x-value of the vertex is the axis. Plug that x into y = ax² + bx + c.

The result is the y-value.

Write the vertex as a point (x, y).

Y-intercept

Read the constant, or substitute x = 0.

When x = 0, every term with x is 0, so y = c.

The y-intercept is the point (0, c).

Example

y = x² + 2x + 6

Substitute x = 0.

y = (0)² + 2(0) + 6

y = 0 + 0 + 6

y = 6

The y-intercept is (0, 6).

Opens up or down

The sign of a tells you which way the parabola opens.

a > 0: the parabola opens up. The vertex is a minimum.

a < 0: the parabola opens down. The vertex is a maximum.

The minimum or maximum value is the y-value of the vertex.

y = x²

a = 1 and a > 0. Opens up. Minimum value: 0.

y = −x²

a = −1 and a < 0. Opens down. Maximum value: 0.

Easy 1

Axis of Symmetry

Your answer must be an equation x = …

Example

Find the axis of symmetry of y = x² − 8x + 3. Your answer must be an equation.

The formula is

x = −b2a

Substitute.

x = −(−8)2(1)

x = 82

Write the answer as an equation.

x = 4

Answer: x = 4

Find the axis of symmetry of y = x² − 4x + 7. Your answer must be an equation.

Show Answer

The formula is

x = −b2a

Substitute.

x = −(−4)2(1)

x = 42

Write the answer as an equation.

x = 2

Answer: x = 2

Find the axis of symmetry of y = x² + 6x − 2. Your answer must be an equation.

Show Answer

The formula is

x = −b2a

Substitute.

x = −62(1)

x = −62

Write the answer as an equation.

x = −3

Answer: x = −3

Find the axis of symmetry of y = x² − 2x + 5. Your answer must be an equation.

Show Answer

The formula is

x = −b2a

Substitute.

x = −(−2)2(1)

x = 22

Write the answer as an equation.

x = 1

Answer: x = 1

Find the axis of symmetry of y = x² + 10x + 1. Your answer must be an equation.

Show Answer

The formula is

x = −b2a

Substitute.

x = −102(1)

x = −102

Write the answer as an equation.

x = −5

Answer: x = −5

Find the axis of symmetry of y = −x² + 8x + 4. Your answer must be an equation.

Show Answer

The formula is

x = −b2a

Substitute.

x = −82(−1)

x = −8−2

x = 82

Write the answer as an equation.

x = 4

Answer: x = 4

Easy 2

Vertex

Find the axis, then plug that x back in. Write the vertex as a point.

Example

Find the vertex of y = x² − 4x + 1.

The formula is

x = −b2a

Substitute.

x = −(−4)2(1)

x = 42

x = 2

Plug this x into the equation.

y = x² − 4x + 1

Substitute.

y = (2)² − 4(2) + 1

y = 4 − 8 + 1

y = −3

Vertex: (2, −3)

Answer: (2, −3)

Find the vertex of y = x² − 6x + 5.

Show Answer

The formula is

x = −b2a

Substitute.

x = −(−6)2(1)

x = 62

x = 3

Plug this x into the equation.

y = x² − 6x + 5

Substitute.

y = (3)² − 6(3) + 5

y = 9 − 18 + 5

y = −4

Vertex: (3, −4)

Answer: (3, −4)

Find the vertex of y = x² + 2x − 8.

Show Answer

The formula is

x = −b2a

Substitute.

x = −22(1)

x = −22

x = −1

Plug this x into the equation.

y = x² + 2x − 8

Substitute.

y = (−1)² + 2(−1) − 8

y = 1 − 2 − 8

y = −9

Vertex: (−1, −9)

Answer: (−1, −9)

Find the vertex of y = x² − 10x + 16.

Show Answer

The formula is

x = −b2a

Substitute.

x = −(−10)2(1)

x = 102

x = 5

Plug this x into the equation.

y = x² − 10x + 16

Substitute.

y = (5)² − 10(5) + 16

y = 25 − 50 + 16

y = −9

Vertex: (5, −9)

Answer: (5, −9)

Find the vertex of y = x² + 8x + 9.

Show Answer

The formula is

x = −b2a

Substitute.

x = −82(1)

x = −82

x = −4

Plug this x into the equation.

y = x² + 8x + 9

Substitute.

y = (−4)² + 8(−4) + 9

y = 16 − 32 + 9

y = −7

Vertex: (−4, −7)

Answer: (−4, −7)

Find the vertex of y = −x² + 6x + 2.

Show Answer

The formula is

x = −b2a

Substitute.

x = −62(−1)

x = −6−2

x = 62

x = 3

Plug this x into the equation.

y = −x² + 6x + 2

Substitute.

y = −(3)² + 6(3) + 2

y = −9 + 18 + 2

y = 11

Vertex: (3, 11)

Answer: (3, 11)

Find the vertex of y = −x² + 4x + 3.

Show Answer

The formula is

x = −b2a

Substitute.

x = −42(−1)

x = −4−2

x = 42

x = 2

Plug this x into the equation.

y = −x² + 4x + 3

Substitute.

y = −(2)² + 4(2) + 3

y = −4 + 8 + 3

y = 7

Vertex: (2, 7)

Answer: (2, 7)

Easy 3

Y-intercept

The y-intercept is the point where x = 0.

Example

Find the y-intercept of y = x² − 4x + 8.

y = x² − 4x + 8

Substitute x = 0.

y = (0)² − 4(0) + 8

y = 0 + 0 + 8

y = 8

The y-intercept is (0, 8).

Answer: (0, 8)

Find the y-intercept of y = x² + 3x − 4.

Show Answer

y = x² + 3x − 4

Substitute x = 0.

y = (0)² + 3(0) − 4

y = 0 + 0 − 4

y = −4

The y-intercept is (0, −4).

Answer: (0, −4)

Find the y-intercept of y = x² − 5x + 2.

Show Answer

y = x² − 5x + 2

Substitute x = 0.

y = (0)² − 5(0) + 2

y = 0 + 0 + 2

y = 2

The y-intercept is (0, 2).

Answer: (0, 2)

Find the y-intercept of y = −x² + 4x + 9.

Show Answer

y = −x² + 4x + 9

Substitute x = 0.

y = −(0)² + 4(0) + 9

y = 0 + 0 + 9

y = 9

The y-intercept is (0, 9).

Answer: (0, 9)

Find the y-intercept of y = x² − 3.

Show Answer

y = x² − 3

Substitute x = 0.

y = (0)² − 3

y = 0 − 3

y = −3

The y-intercept is (0, −3).

Answer: (0, −3)

Find the y-intercept of y = 2x² − x + 7.

Show Answer

y = 2x² − x + 7

Substitute x = 0.

y = 2(0)² − (0) + 7

y = 0 + 0 + 7

y = 7

The y-intercept is (0, 7).

Answer: (0, 7)

Find the y-intercept of y = 4x² + 3x.

Show Answer

y = 4x² + 3x

Substitute x = 0.

y = 4(0)² + 3(0)

y = 0 + 0

y = 0

The y-intercept is (0, 0).

Answer: (0, 0)

Easy 4

Minimum and Maximum

You are given the vertex. Use the sign of a. The value is the y-value only.

Example

Given y = x² − 2x + 6. The vertex is (1, 5). Using whether the parabola opens up or down, what is the minimum or maximum value?

y = x² − 2x + 6

Vertex: (1, 5)

a = 1, so a > 0.

The parabola opens up.

The vertex is a minimum.

The minimum value is the y-value of the vertex.

Minimum: 5

Answer: Minimum: 5

Given y = x² + 4x + 1. The vertex is (−2, −3). Using whether the parabola opens up or down, what is the minimum or maximum value?

Show Answer

y = x² + 4x + 1

Vertex: (−2, −3)

a = 1, so a > 0.

The parabola opens up.

The vertex is a minimum.

The minimum value is the y-value of the vertex.

Minimum: −3

Answer: Minimum: −3

Given y = −x² + 2x + 4. The vertex is (1, 5). Using whether the parabola opens up or down, what is the minimum or maximum value?

Show Answer

y = −x² + 2x + 4

Vertex: (1, 5)

a = −1, so a < 0.

The parabola opens down.

The vertex is a maximum.

The maximum value is the y-value of the vertex.

Maximum: 5

Answer: Maximum: 5

Given y = x² − 6x + 2. The vertex is (3, −7). Using whether the parabola opens up or down, what is the minimum or maximum value?

Show Answer

y = x² − 6x + 2

Vertex: (3, −7)

a = 1, so a > 0.

The parabola opens up.

The vertex is a minimum.

The minimum value is the y-value of the vertex.

Minimum: −7

Answer: Minimum: −7

Given y = −x² − 6x − 4. The vertex is (−3, 5). Using whether the parabola opens up or down, what is the minimum or maximum value?

Show Answer

y = −x² − 6x − 4

Vertex: (−3, 5)

a = −1, so a < 0.

The parabola opens down.

The vertex is a maximum.

The maximum value is the y-value of the vertex.

Maximum: 5

Answer: Maximum: 5

Given y = 2x² + 4x + 5. The vertex is (−1, 3). Using whether the parabola opens up or down, what is the minimum or maximum value?

Show Answer

y = 2x² + 4x + 5

Vertex: (−1, 3)

a = 2, so a > 0.

The parabola opens up.

The vertex is a minimum.

The minimum value is the y-value of the vertex.

Minimum: 3

Answer: Minimum: 3

Easy 5

Domain and Range

The vertex is labeled on the graph. Write the domain and range in interval notation.

Example

State the domain and range in interval notation.

Domain: the graph goes left and right forever.

(−∞, ∞)

The vertex on the graph is (1, 3).

The parabola opens down.

Range: (−∞, 3]

Answer: Domain: (−∞, ∞)Range: (−∞, 3]

State the domain and range in interval notation.

Show Answer

Domain: the graph goes left and right forever.

(−∞, ∞)

The vertex on the graph is (2, 0).

The parabola opens up.

Range: [0, ∞)

Answer: Domain: (−∞, ∞)Range: [0, ∞)

State the domain and range in interval notation.

Show Answer

Domain: the graph goes left and right forever.

(−∞, ∞)

The vertex on the graph is (3, 4).

The parabola opens down.

Range: (−∞, 4]

Answer: Domain: (−∞, ∞)Range: (−∞, 4]

State the domain and range in interval notation.

Show Answer

Domain: the graph goes left and right forever.

(−∞, ∞)

The vertex on the graph is (−1, −3).

The parabola opens up.

Range: [−3, ∞)

Answer: Domain: (−∞, ∞)Range: [−3, ∞)

State the domain and range in interval notation.

Show Answer

Domain: the graph goes left and right forever.

(−∞, ∞)

The vertex on the graph is (−1, 3).

The parabola opens down.

Range: (−∞, 3]

Answer: Domain: (−∞, ∞)Range: (−∞, 3]

State the domain and range in interval notation.

Show Answer

Domain: the graph goes left and right forever.

(−∞, ∞)

The vertex on the graph is (3, 2).

The parabola opens up.

Range: [2, ∞)

Answer: Domain: (−∞, ∞)Range: [2, ∞)

Easy 6

Table of 5 Points

The vertex and the y-intercept are filled in. Choose an x-value and substitute it into the equation to find the partner y. Your three points may be different.

Example

Make a table of 5 points for y = x² − 2x − 3. The vertex and the y-intercept are filled in. Choose three other x-values and substitute each one to find its partner y. Your points may be different.

x	y
1	−4	vertex
0	−3	y-intercept

Choose an x-value. Substitute it into the equation. That result is the partner y.

The vertex and the y-intercept are already in the table.

Choose x = −1.

y = x² − 2x − 3

Substitute.

y = (−1)² − 2(−1) − 3

y = 1 + 2 − 3

y = 0

Point: (−1, 0)

Choose x = 2.

y = x² − 2x − 3

Substitute.

y = (2)² − 2(2) − 3

y = 4 − 4 − 3

y = −3

Point: (2, −3)

Choose x = 3.

y = x² − 2x − 3

Substitute.

y = (3)² − 2(3) − 3

y = 9 − 6 − 3

y = 0

Point: (3, 0)

One possible table:

x	y
1	−4	vertex
0	−3	y-intercept
−1	0
2	−3
3	0

Your three points may be different if you chose different x-values.

Answer: One possible set: (1, −4), (0, −3), (−1, 0), (2, −3), (3, 0).Your points may be different if you chose different x-values.

Make a table of 5 points for y = x² − 6x + 8. The vertex and the y-intercept are filled in. Choose three other x-values and substitute each one to find its partner y. Your points may be different.

x	y
3	−1	vertex
0	8	y-intercept

Show Answer

Choose an x-value. Substitute it into the equation. That result is the partner y.

The vertex and the y-intercept are already in the table.

Choose x = 2.

y = x² − 6x + 8

Substitute.

y = (2)² − 6(2) + 8

y = 4 − 12 + 8

y = 0

Point: (2, 0)

Choose x = 4.

y = x² − 6x + 8

Substitute.

y = (4)² − 6(4) + 8

y = 16 − 24 + 8

y = 0

Point: (4, 0)

Choose x = 6.

y = x² − 6x + 8

Substitute.

y = (6)² − 6(6) + 8

y = 36 − 36 + 8

y = 8

Point: (6, 8)

One possible table:

x	y
3	−1	vertex
0	8	y-intercept
2	0
4	0
6	8

Your three points may be different if you chose different x-values.

Answer: One possible set: (3, −1), (0, 8), (2, 0), (4, 0), (6, 8).Your points may be different if you chose different x-values.

Make a table of 5 points for y = x² + 4x + 3. The vertex and the y-intercept are filled in. Choose three other x-values and substitute each one to find its partner y. Your points may be different.

x	y
−2	−1	vertex
0	3	y-intercept

Show Answer

Choose an x-value. Substitute it into the equation. That result is the partner y.

The vertex and the y-intercept are already in the table.

Choose x = −4.

y = x² + 4x + 3

Substitute.

y = (−4)² + 4(−4) + 3

y = 16 − 16 + 3

y = 3

Point: (−4, 3)

Choose x = −3.

y = x² + 4x + 3

Substitute.

y = (−3)² + 4(−3) + 3

y = 9 − 12 + 3

y = 0

Point: (−3, 0)

Choose x = −1.

y = x² + 4x + 3

Substitute.

y = (−1)² + 4(−1) + 3

y = 1 − 4 + 3

y = 0

Point: (−1, 0)

One possible table:

x	y
−2	−1	vertex
0	3	y-intercept
−4	3
−3	0
−1	0

Your three points may be different if you chose different x-values.

Answer: One possible set: (−2, −1), (0, 3), (−4, 3), (−3, 0), (−1, 0).Your points may be different if you chose different x-values.

Make a table of 5 points for y = −x² + 4x + 1. The vertex and the y-intercept are filled in. Choose three other x-values and substitute each one to find its partner y. Your points may be different.

x	y
2	5	vertex
0	1	y-intercept

Show Answer

Choose an x-value. Substitute it into the equation. That result is the partner y.

The vertex and the y-intercept are already in the table.

Choose x = 1.

y = −x² + 4x + 1

Substitute.

y = −(1)² + 4(1) + 1

y = −1 + 4 + 1

y = 4

Point: (1, 4)

Choose x = 3.

y = −x² + 4x + 1

Substitute.

y = −(3)² + 4(3) + 1

y = −9 + 12 + 1

y = 4

Point: (3, 4)

Choose x = 4.

y = −x² + 4x + 1

Substitute.

y = −(4)² + 4(4) + 1

y = −16 + 16 + 1

y = 1

Point: (4, 1)

One possible table:

x	y
2	5	vertex
0	1	y-intercept
1	4
3	4
4	1

Your three points may be different if you chose different x-values.

Answer: One possible set: (2, 5), (0, 1), (1, 4), (3, 4), (4, 1).Your points may be different if you chose different x-values.

Make a table of 5 points for y = 2x² − 8x + 3. The vertex and the y-intercept are filled in. Choose three other x-values and substitute each one to find its partner y. Your points may be different.

x	y
2	−5	vertex
0	3	y-intercept

Show Answer

Choose an x-value. Substitute it into the equation. That result is the partner y.

The vertex and the y-intercept are already in the table.

Choose x = 1.

y = 2x² − 8x + 3

Substitute.

y = 2(1)² − 8(1) + 3

y = 2 − 8 + 3

y = −3

Point: (1, −3)

Choose x = 3.

y = 2x² − 8x + 3

Substitute.

y = 2(3)² − 8(3) + 3

y = 18 − 24 + 3

y = −3

Point: (3, −3)

Choose x = 4.

y = 2x² − 8x + 3

Substitute.

y = 2(4)² − 8(4) + 3

y = 32 − 32 + 3

y = 3

Point: (4, 3)

One possible table:

x	y
2	−5	vertex
0	3	y-intercept
1	−3
3	−3
4	3

Your three points may be different if you chose different x-values.

Answer: One possible set: (2, −5), (0, 3), (1, −3), (3, −3), (4, 3).Your points may be different if you chose different x-values.

Make a table of 5 points for y = −x² − 2x + 3. The vertex and the y-intercept are filled in. Choose three other x-values and substitute each one to find its partner y. Your points may be different.

x	y
−1	4	vertex
0	3	y-intercept

Show Answer

Choose an x-value. Substitute it into the equation. That result is the partner y.

The vertex and the y-intercept are already in the table.

Choose x = −3.

y = −x² − 2x + 3

Substitute.

y = −(−3)² − 2(−3) + 3

y = −9 + 6 + 3

y = 0

Point: (−3, 0)

Choose x = −2.

y = −x² − 2x + 3

Substitute.

y = −(−2)² − 2(−2) + 3

y = −4 + 4 + 3

y = 3

Point: (−2, 3)

Choose x = 1.

y = −x² − 2x + 3

Substitute.

y = −(1)² − 2(1) + 3

y = −1 − 2 + 3

y = 0

Point: (1, 0)

One possible table:

x	y
−1	4	vertex
0	3	y-intercept
−3	0
−2	3
1	0

Your three points may be different if you chose different x-values.

Answer: One possible set: (−1, 4), (0, 3), (−3, 0), (−2, 3), (1, 0).Your points may be different if you chose different x-values.

Medium

Graph and Find Everything

Sketch the parabola. List the axis, vertex, minimum or maximum, y-intercept, domain, and range.

Example

Given y = 2x² + 8x − 3. Sketch the graph and list the following. Give the domain and range in interval notation.a. Axis of symmetryb. Vertexc. Minimum or maximumd. Y-intercepte. Domainf. Range

The formula is

x = −b2a

Substitute.

x = −84

Write the axis as an equation.

x = −2

Vertex: plug this x into the equation.

y = 2x² + 8x − 3

Substitute.

y = 2(−2)² + 8(−2) − 3

y = −11

(−2, −11)

a > 0, so the parabola opens up. The vertex is a minimum.

The minimum value is the y-value of the vertex.

Minimum: −11

Y-intercept: x = 0.

(0, −3)

Domain: all real x.

(−∞, ∞)

Range.

[−11, ∞)

Sketch.

Answer: a. Axis of Symmetry: x = −2b. Vertex: (−2, −11)c. Minimum: −11d. Y-intercept: (0, −3)e. Domain: (−∞, ∞)f. Range: [−11, ∞)

Given y = 2x² − 12x + 7. Sketch the graph and list the following. Give the domain and range in interval notation.a. Axis of symmetryb. Vertexc. Minimum or maximumd. Y-intercepte. Domainf. Range

Show Answer

The formula is

x = −b2a

Substitute.

x = 124

Write the axis as an equation.

x = 3

Vertex: plug this x into the equation.

y = 2x² − 12x + 7

Substitute.

y = 2(3)² − 12(3) + 7

y = −11

(3, −11)

a > 0, so the parabola opens up. The vertex is a minimum.

The minimum value is the y-value of the vertex.

Minimum: −11

Y-intercept: x = 0.

(0, 7)

Domain: all real x.

(−∞, ∞)

Range.

[−11, ∞)

Sketch.

Answer: a. Axis of Symmetry: x = 3b. Vertex: (3, −11)c. Minimum: −11d. Y-intercept: (0, 7)e. Domain: (−∞, ∞)f. Range: [−11, ∞)

Given y = −2x² − 4x + 3. Sketch the graph and list the following. Give the domain and range in interval notation.a. Axis of symmetryb. Vertexc. Minimum or maximumd. Y-intercepte. Domainf. Range

Show Answer

The formula is

x = −b2a

Substitute.

x = 4−4

Write the axis as an equation.

x = −1

Vertex: plug this x into the equation.

y = −2x² − 4x + 3

Substitute.

y = −2(−1)² − 4(−1) + 3

y = 5

(−1, 5)

a < 0, so the parabola opens down. The vertex is a maximum.

The maximum value is the y-value of the vertex.

Maximum: 5

Y-intercept: x = 0.

(0, 3)

Domain: all real x.

(−∞, ∞)

Range.

(−∞, 5]

Sketch.

Answer: a. Axis of Symmetry: x = −1b. Vertex: (−1, 5)c. Maximum: 5d. Y-intercept: (0, 3)e. Domain: (−∞, ∞)f. Range: (−∞, 5]

Given y = 3x² + 6x − 4. Sketch the graph and list the following. Give the domain and range in interval notation.a. Axis of symmetryb. Vertexc. Minimum or maximumd. Y-intercepte. Domainf. Range

Show Answer

The formula is

x = −b2a

Substitute.

x = −66

Write the axis as an equation.

x = −1

Vertex: plug this x into the equation.

y = 3x² + 6x − 4

Substitute.

y = 3(−1)² + 6(−1) − 4

y = −7

(−1, −7)

a > 0, so the parabola opens up. The vertex is a minimum.

The minimum value is the y-value of the vertex.

Minimum: −7

Y-intercept: x = 0.

(0, −4)

Domain: all real x.

(−∞, ∞)

Range.

[−7, ∞)

Sketch.

Answer: a. Axis of Symmetry: x = −1b. Vertex: (−1, −7)c. Minimum: −7d. Y-intercept: (0, −4)e. Domain: (−∞, ∞)f. Range: [−7, ∞)

Given y = −3x² + 12x − 5. Sketch the graph and list the following. Give the domain and range in interval notation.a. Axis of symmetryb. Vertexc. Minimum or maximumd. Y-intercepte. Domainf. Range

Show Answer

The formula is

x = −b2a

Substitute.

x = −12−6

Write the axis as an equation.

x = 2

Vertex: plug this x into the equation.

y = −3x² + 12x − 5

Substitute.

y = −3(2)² + 12(2) − 5

y = 7

(2, 7)

a < 0, so the parabola opens down. The vertex is a maximum.

The maximum value is the y-value of the vertex.

Maximum: 7

Y-intercept: x = 0.

(0, −5)

Domain: all real x.

(−∞, ∞)

Range.

(−∞, 7]

Sketch.

Answer: a. Axis of Symmetry: x = 2b. Vertex: (2, 7)c. Maximum: 7d. Y-intercept: (0, −5)e. Domain: (−∞, ∞)f. Range: (−∞, 7]

Given y = 2x² + 4x + 1. Sketch the graph and list the following. Give the domain and range in interval notation.a. Axis of symmetryb. Vertexc. Minimum or maximumd. Y-intercepte. Domainf. Range

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The formula is

x = −b2a

Substitute.

x = −44

Write the axis as an equation.

x = −1

Vertex: plug this x into the equation.

y = 2x² + 4x + 1

Substitute.

y = 2(−1)² + 4(−1) + 1

y = −1

(−1, −1)

a > 0, so the parabola opens up. The vertex is a minimum.

The minimum value is the y-value of the vertex.

Minimum: −1

Y-intercept: x = 0.

(0, 1)

Domain: all real x.

(−∞, ∞)

Range.

[−1, ∞)

Sketch.

Answer: a. Axis of Symmetry: x = −1b. Vertex: (−1, −1)c. Minimum: −1d. Y-intercept: (0, 1)e. Domain: (−∞, ∞)f. Range: [−1, ∞)

Hard

Fractional Vertex

The axis is a fraction. Find the vertex and write the point.

Example

Find the vertex of y = x² + 3x + 1.

The formula is

x = −b2a

Substitute.

x = −32(1)

x = −32

Plug this x into the equation.

y = x² + 3x + 1

Substitute.

y = (−32)² + 3(−32) + 1

(−32)² = 94

3 · (−32) = −92

y = 94 − 92 + 1

y = 94 − 184 + 44

Add the numerators.

9 − 18 + 4 = −5

y = −54

Vertex: (−32, −54)

Answer: (−32, −54)

Find the vertex of y = x² − x + 4.

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The formula is

x = −b2a

Substitute.

x = −(−1)2(1)

x = 12

Plug this x into the equation.

y = x² − x + 4

Substitute.

y = (12)² − (12) + 4

(12)² = 14

y = 14 − 12 + 4

y = 14 − 24 + 164

Add the numerators.

1 − 2 + 16 = 15

y = 154

Vertex: (12, 154)

Answer: (12, 154)

Find the vertex of y = −2x² + 6x + 1.

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The formula is

x = −b2a

Substitute.

x = −62(−2)

x = −6−4

x = 64

x = 32

Plug this x into the equation.

y = −2x² + 6x + 1

Substitute.

y = −2(32)² + 6(32) + 1

(32)² = 94

−2 · 94 = −92

6 · (32) = 9

y = −92 + 9 + 1

y = −92 + 182 + 22

Add the numerators.

−9 + 18 + 2 = 11

y = 112

Vertex: (32, 112)

Answer: (32, 112)

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