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

Question

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

0

Answers ( No )

  1. Charlotte
    0
    2021-11-26T07:58:14+00:00

    Answer:

    A

    Step-by-step explanation:

  2. Charlotte
    0
    2021-11-26T07:58:22+00:00

    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

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