HOW TO FIND THE VERTEX OF A QUADRATIC FUNCTION IN GENERAL FORM

In each case find the vertex.

Example 1 :

y = x² + 6x + 7

Solution :

Comparing y = ax² + bx + c and y = x² + 6x + 7, we get

a = 1  and  b = 6

x-coordinate of the vertex :

y-coordinate of the vertex :

Substitute x = -3 into the given quadratic function.

y = (-3)² + 6(-3) + 7

y = 9 - 18 + 7

y = -2

Vertex :

(-3, -2)

Example 2 :

y = 2x² - 8x + 5

Solution :

Comparing y = ax² + bx + c and y = 2x² - 8x + 5, we get

a = 2  and  b = -8

x-coordinate of the vertex :

y-coordinate of the vertex :

Substitute x = 2 into the given quadratic function.

y = 2(2)² - 8(2) + 5

y = 2(4) - 16 + 5

y = 8 - 16 + 5

y = -3

Vertex :

(2, -3)

Example 3 :

y = x² - 5x - 3

Solution :

Comparing y = ax² + bx + c and y = x² - 5x - 3, we get

a = 1  and  b = -5

x-coordinate of the vertex :

y-coordinate of the vertex :

Substitute x = 2.5 into the given quadratic function.

y = (2.5)² - 5(2.5) - 3

y = 6.25 - 12.5 + 5

y = -1.25

Vertex :

(2.5, -1.25)

Example 4 :

x = y² + 4y - 5

Solution :

Comparing x = ay² + by + c and x = y² + 4y - 5, we get

a = 1  and  b = 4

y-coordinate of the vertex :

x-coordinate of the vertex :

Substitute y = -2 into the given quadratic function.

x = (-2)² + 4(-2) - 5

x = 4 - 8 - 5

x = -9

Vertex :

(-9, -2)

Example 5 :

x = 2y² - 8y - 7

Solution :

Comparing x = ay² + by + c and x = 2y² - 8y - 7, we get

a = 2  and  b = -8

y-coordinate of the vertex :

x-coordinate of the vertex :

Substitute y = 2 into the given quadratic function.

x = 2(2)² - 8(2) - 7

x = 2(4) - 16 - 7

x = 8 - 16 - 7

x = -15

Vertex :

(-15, 2)

Example 6 :

x = y² + 7y + 1

Solution :

Comparing x = ay² + by + c and x = y² + 7y + 1, we get

a = 1  and  b = 7

y-coordinate of the vertex :

x-coordinate of the vertex :

Substitute y = -3.5 into the given quadratic function.

x = (-3.5)² + 7(-3.5) + 1

x = 12.25 - 24.5 + 1

x = -11.25

Vertex :

(-11.25, -3.5)

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