Please help me. This is due tonight.

1. vertex is at point (-3,5)

2.vertex is at point (-1,-6)

3.vertex is at point (3,0)

Step-by-step explanation:

Use y = ax^2+bx+c

Use x = -b/2a

1.

y = 3x^2 + 18x + 32

identify a, b and c

a = 3, b = 18, c = 32

Use x = -b/2a

x = -18/2(3) = -3

substitute back into y = 3x^2 + 18x + 32

y = 3(-3)^2 + 18(-3) + 32 = 5

vertex is at point (-3,5)

2.

y = x^2 + 2x - 5

identify a, b and c

a = 1, b = 2, c = -5

Use x = -b/2a

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

substitute back into y = x^2 + 2x - 5

y = (-1)^2 + 2(-1) - 5 = -6

vertex is at point (-1,-6)

3.

y = -3x^2 + 18x - 27

identify a, b and c

a = -3, b = 18, c = -27

Use x = -b/2a

x = -18/2(-3) = 3

substitute back into y = -3x^2 + 18x - 27

y = -3(3)^2 + 18(3) - 27 = 0

vertex is at point (3,0)

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.

2

step-by-step explanation: