## MATH PLEASE HELP! Identify the vertex and the y-intercept of Y=-2(x+3)^2+2the graph of the function . A.vertex: (−3, 2); y-i

MATH PLEASE HELP!

Identify the vertex and the y-intercept of Y=-2(x+3)^2+2the graph of the function .

A.vertex: (−3, 2); y-intercept: −16

B.vertex: (−3, −2); y-intercept: 11

C.vertex: (3, 2); y-intercept: −16

D. vertex: (3, −2); y-intercept: −18

Identify the vertex and the y-intercept of the graph of the function Y=-2(x+2)^2+2

A.vertex: (−2, 2); y-intercept: −6

B.vertex: (2, 2); y-intercept: −6

C. vertex: (−2, −2); y-intercept: 6

D. vertex: (2, −2); y-intercept: −8

## Answers ( No )

Answer:A

Step-by-step explanation:A and A

the equation of a parabola in vertex form is

y = a(x – h)² + k

where ( h, k ) are the coordinates of the vertex and a is a multiplier

y = – 2(x + 3)² + 2 is in this form

with vertex = ( – 3, 2)

To find the y-intercept let x = 0

y = – 2(3)² + 2 = – 18 + 2 = – 16

Similarly

y = – 2(x + 2)² + 2 is in vertex form

vertex = ( – 2 , 2)

x = 0 : y = – 2(2)² + 2 = – 8 + 2 = – 6 ← y- intercept