Respuesta :

TREKKER

The vertex of any equation in the form y = a(x-b)^2 + c is (b,c). Keeping that in mind, we figure out that:

The first one is (-9,-5)

The second one is (-5,-6)

The third one is (5,6)

The fourth one is (6,9)

The fifth one is (5,-9)

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Answer:

see below

Step-by-step explanation:

The vertex form of a parabola is given by

y = a(x-h)^2 +k

y = 6(x+9)^2 -5 =   6(x- -9)^2 +-5  The vertex is (-9,-5)

y = 9(x+5)^2 -6 =   9(x- -5)^2 +-6  The vertex is (-5,-6)

y = 9(x-5)^2 +6 =                            The vertex is (5,6)

y = 5(x-6)^2 +9 =                           The vertex is ( 6,9)

y = 6(x-5)^2 -9 =   6(x -5)^2 + -9  The vertex is (5,-9)

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