Respuesta :

wegnerkolmp2741o

Answer:

see below

Step-by-step explanation:

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

Where (h,k) is the vertex

f(x) = 9(x-5)^2 + 6.  has a vertex of (5,6)

f(x) = 6(x- -9)^2  + -5  has a vertex of (-9, -5)

f(x) = 9(x- -5)^2  + -6  has a vertex of (-5, -6)

f(x) = 5(x- 6)^2  + 9  has a vertex of (6, 9)

f(x) = 6(x- 5)^2  + -9  has a vertex of (5, -9)

amna04352

Answer:

1) (5,6)

2) (-9,-5)

3) (-5,-6)

4) (6,9)

5) (5,-9)

Step-by-step explanation:

In a(x- h)² + k

Vertex is (h,k)

For h:

Equate the square bracket to 0

(x - h) = 0

x = h

For k:

Just pick the constant outside the square

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