The vertex form of a quadratic function is f(x) = a(x-h)2 + k. What is the vertex of each function?

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)

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

the answer is somewhere there guys it really . there im just not gonna answer

step-by-step explanation:

but i'm guessing ur first option tho

the answer is b: 18

step-by-step explanation:

use the pythagorean theorem to .

a^2+b^2=c^2

plug 18 for "r"

the 2 legs are 18 and 24

18^2+24^2=c^2

324+576=c^2

900=c^2

c=30

30-12=18

18 is the radius

: )