FINDING VERTEX OF A PARABOLA IN STANDARD FORM

Name a, b and c for each parabola. Then find the vertex. Show all work.

Problem 1 :

f(x) = x² - 8x + 17

Solution :

Comparing f(x) = ax² + bx + c and f(x) = x² - 8x + 17

a = 1, b = -8 and c = 17

x-coordinate of the vertex :

x = -b/2a

Substitute a = 1 and b = -8.

x = -(-8)/2(1)

x = 8/2

x = 4

y-coordinate of the vertex :

Substitute x = 4 in f(x) = x² - 8x + 17

f(4) = 4² - 8(4) + 17

= 16 - 32 + 17

= 1

Vertex of the parabola is (4, 1).

Problem 2 :

f(x) = -x² - 2x - 2

Solution : 

Comparing f(x) = ax² + bx + c and f(x) = -x² - 2x - 2

a = -1, b = -2 and c = -2

x-coordinate of the vertex :

x = -b/2a

Substitute a = -1 and b = -2.

x = -(-2)/2(-1)

x = 2/-2

x = -1

y-coordinate of the vertex :

Substitute x = -1 in f(x) = -x² - 2x - 2

f(-1) = -(-1)² - 2(-1) - 2

= -1 + 2 - 2

= 1

Vertex of the parabola is (-1, 1).

Problem 3 :

f(x) = -x² + 6x - 8

Solution :

Comparing f(x) = ax² + bx + c and f(x) = -x² + 6x - 8

a = -1, b = 6 and c = -8

x - coordinate of the vertex :

x = -b/2a

Substitute a = -1 and b = 6.

x = -6/2(-1)

x = -6/-2

x = 3

y - coordinate of the vertex :

Substitute x = 3 in f(x) = -x² + 6x - 8

f(3) = -3² + 6(3) - 8

= -9 + 18 - 8

= 1

Vertex of the parabola is (3, 1).

Problem 4 :

f(x) = -3x² + 6x

Solution : 

Comparing f(x) = ax² + bx + c and f(x) = -3x² + 6x

a = -3, b = 6 and c = 0

x-coordinate of the vertex :

x = -b/2a

Substitute a = -3 and b = 6.

x = -6/2(-3)

x = -6/-6

x = 1

y-coordinate of the vertex :

Substitute x = 1 in f(x) = -3x² + 6x

f(1) = -3(1)² + 6(1)

= -3 + 6

= 3

Vertex of the parabola is (1, 3).

Problem 5 :

f(x) = -2x² - 16x - 31

Solution :

Comparing f(x) = ax² + bx + c and f(x) = -2x² - 16x - 31

a = -2, b = -16 and c = -31

x-coordinate of the vertex :

x = -b/2a

Substitute a = -2 and b = -16.

x = -(-16)/2(-2)

x = 16/-4

x = -4

y-coordinate of the vertex :

Substitute x = -4 in f(x) = -2x² - 16x - 31

f(-4) = -2(-4)² - 16(-4) - 31

= -32 + 64 - 31

= 64 - 63

= 1

Vertex of the parabola is (-4, 1).

Problem 6 :

f(x) = -1/2x² - 4x - 6

Solution :

Comparing f(x) = ax² + bx + c and f(x) = -1/2x² - 4x - 6

a = -1/2, b = -4 and c = -6

x-coordinate of the vertex :

x = -b/2a

Substitute a = -1/2 and b = -4.

x = -(-4)/2(-1/2)

x = 4/-1

x = -4

y-coordinate of the vertex :

Substitute x = -4 in f(x) = -1/2x² - 4x - 6

f(-4) = -1/2(-4)² - 4(-4) - 6

= -8 + 16 - 6

= 16 - 14

= 2

Vertex of the parabola is (-4, 2).

Problem 7 :

f(x) = -2x² + 12x - 22

Solution :

Comparing f(x) = ax² + bx + c and f(x) = -2x² + 12x - 22

a = -2, b = 12 and c = -22

x-coordinate of the vertex :

x = -b/2a

Substitute a = -2 and b = 12.

x = -12/2(-2)

x = -12/-4

x = 3

y-coordinate of the vertex :

Substitute x = 3 in f(x) = -2x² + 12x - 22

f(3) = -2(3)² + 12(3) - 22

= -18 + 36 - 22

= -40 + 36

= -4

Vertex of the parabola is (3, -4).

Problem 8 :

f(x) = x² - 8x + 20

Solution :

Comparing f(x) = ax² + bx + c and f(x) = x² - 8x + 20

a = 1, b = -8 and c = 20

x-coordinate of the vertex:

x = -b/2a

Substitute a = 1 and b = -8.

x = -(-8)/2(1)

x = 8/2

x = 4

y-coordinate of the vertex:

Substitute x = 4 in f(x) = x² - 8x + 20

f(4) = 4² - 8(4) + 20

= 16 - 32 + 20

= 4

Vertex of the parabola is (4, 4).