MATH PLEASE HELP!
Identify the vertex and the y-intercept of Y=-2(x+3)^2+2the graph of the function .
A.vertex: (βˆ’3, 2); y-intercept: βˆ’16
B.vertex: (βˆ’3, βˆ’2); y-intercept: 11
C.vertex: (3, 2); y-intercept: βˆ’16
D. vertex: (3, βˆ’2); y-intercept: βˆ’18

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

A.vertex: (βˆ’2, 2); y-intercept: βˆ’6


B.vertex: (2, 2); y-intercept: βˆ’6


C. vertex: (βˆ’2, βˆ’2); y-intercept: 6


D. vertex: (2, βˆ’2); y-intercept: βˆ’8



Respuesta :

A and A

the equation of a parabola in vertex form is

y = a(x - h)Β² + k

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

y = - 2(x + 3)Β² + 2 Β is in this form

with vertex = ( - 3, 2)

To find the y-intercept let x = 0

y = - 2(3)Β² + 2 = - 18 + 2 = - 16

Similarly

y = - 2(x + 2)Β² + 2 is in vertex form

vertex = ( - 2 , 2)

x = 0 : y = - 2(2)Β² + 2 = - 8 + 2 = - 6 ← y- intercept


Answer:

A

Step-by-step explanation:

good job