# QUADRATIC FUNCTION AXIS OF SYMMETRY AND VERTEX WORKSHEET

Find the axis of symmetry and vertex for the following quadratic functions.

Problem 1 :

y = x2 - 2x - 35

Problem 2 :

y = -2x2 + 6x - 3

Problem 3 :

y = 2x2 - 16x + 3

Problem 4 :

y = -x2 - 4x + 1

Problem 5 :

y = 3x2 - 6x + 4

Problem 6 :

y = 3(x - 1)2 + 3

Problem 7 :

y = 2(x - 7)2 - 2

Problem 8 :

y = -2(x + 2)2 + 5

Problem 9 :

y = (x + 3)2 - 2

Problem 10 :

y = (x - 4)2

y = x2 - 2x - 35

The given quadratic function is in standard form.

Comparing y = ax2 + bx + c and y = x2 - 2x - 35, we get

a = 1, b = -2 and  c = -35

Axis of symmetry :

x = 1

From the axis of symmetry x = 1, the x-coordinate of the vertex is 1.

In the given quadratic function, substitute x = 1 to get the y-coordinate of the vertex.

y = 12 - 2(1) - 35

y = 1 - 2 - 35

y = -36

Vertex :

(1, -36)

y = -2x2 + 6x - 3

The given quadratic function is in standard form.

Comparing y = ax2 + bx + c and y = -2x2 + 6x - 3, we get

a = -2, b = 6 and  c = -3

Axis of symmetry :

From the axis of symmetry x = ³⁄₂, the x-coordinate of the vertex is ³⁄₂.

In the given quadratic function, substitute x = ³⁄₂ to get the y-coordinate of the vertex.

Vertex :

y = 2x2 - 16x + 3

The given quadratic function is in standard form.

Comparing y = ax2 + bx + c and y = 2x2 - 16x + 3, we get

a = 2, b = -16 and  c = 3

Axis of symmetry :

x = 4

From the axis of symmetry x = 4, the x-coordinate of the vertex is 4.

In the given quadratic function, substitute x = 4 to get the y-coordinate of the vertex.

y = 2(4)2 - 16(4) + 3

y = 2(16) - 64 + 3

y = 32 - 64 + 3

y = -29

Vertex :

(4, -29)

y = -x2 - 4x + 1

The given quadratic function is in standard form.

Comparing y = ax2 + bx + c and y = -x2 - 4x + 1, we get

a = -1, b = -4 and  c = 1

Axis of symmetry :

x = 2

From the axis of symmetry x = 2, the x-coordinate of the vertex is 2.

In the given quadratic function, substitute x = 2 to get the y-coordinate of the vertex.

y = -(2)2 - 4(2) + 1

y = -4 - 8 + 3

y = -9

Vertex :

(2, -9)

y = 3x2 - 6x + 4

The given quadratic function is in standard form.

Comparing y = ax2 + bx + c and y = 3x2 - 6x + 4, we get

a = 3, b = -6 and  c = 4

Axis of symmetry :

x = 1

From the axis of symmetry x = 1, the x-coordinate of the vertex is 1.

In the given quadratic function, substitute x = 1 to get the y-coordinate of the vertex.

y = 3(1)2 - 6(1) + 4

y = 3(1) - 6 + 4

y = 3 - 6 + 4

y = 1

Vertex :

(1, 1)

y = 3(x - 1)2 + 3

The given quadratic function is in vertex form.

Comparing y = a(x - h)2 + k and y = 3(x - 1)2 + 3, we get

h = 1 and k = 3

Axis of symmetry :

x = h

x = 1

Vertex :

(h, k) = (1, 3)

y = 2(x - 7)2 - 2

y = 2(x - 7)2 + (-2)

The above quadratic function is in vertex form.

Comparing y = a(x - h)2 + k and y = 2(x - 7)2 + (-2), we get

h = 7 and k = -2

Axis of symmetry :

x = h

x = 7

Vertex :

(h, k) = (7, -2)

y = -2(x + 2)2 + 5

y = -2[x - (-2)]2 + 5

The above quadratic function is in vertex form.

Comparing y = a(x - h)2 + k and y = -2[x - (-2)]2 + 5, we get

h = -2 and k = 5

Axis of symmetry :

x = h

x = -2

Vertex :

(h, k) = (-2, 5)

y = (x + 3)2 - 2

y = [x - (-3)]2 + (-2)

The above quadratic function is in vertex form.

Comparing y = a(x - h)2 + k and y = [x - (-3)]2 + (-2), we get

h = -3 and k = -2

Axis of symmetry :

x = h

x = -3

Vertex :

(h, k) = (-3, -2)

y = (x - 4)2

y = (x - 4)+ 0

The above quadratic function is in vertex form.

Comparing y = a(x - h)2 + k and y = (x - 4)+ 0, we get

h = 4 and k = 0

Axis of symmetry :

x = h

x = 4

Vertex :

(h, k) = (4, 0)

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