Which function has a vertex at (2, -9)? O f(x) = -(x – 3)2
O f(x) = (x + 8)2
O f(x) = (x – 5)(x + 1)
O f(x) = -(x - 1)(x – 5)

Explanation:

Lets go through the answer choices to see which have the vertex we're after.

A.

f(x) = -(x-3)^2

y = -1(x-3)^2 + 0

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

a = -1, h = -3, k = 0

The vertex here is (h,k) = (-3,0). We can cross choice A off the list.

B.

f(x) = (x+8)^2

y = 1(x-(-8))^2 + 0

We see the vertex here is (h,k) = (-8,0)

Choice B isn't the answer either.

C.

Find the roots of this given function

f(x) = (x-5)(x+1)

(x-5)(x+1) = 0

x-5 = 0 or x+1 = 0

x = 5 or x = -1

Add these values up and divide by 2

(5+(-1))/2 = 4/2 = 2

The axis of symmetry is x = 2

The axis of symmetry is always the midpoint of the two roots

Plugging x = 2 into f(x) leads to

f(x) = (x-5)(x+1)

f(2) = (2-5)(2+1)

f(2) = (-3)*(3)

f(2) = -9

This shows (2,-9) is the vertex. We can stop here.

If you want to keep going, and check choice D, then let's find the roots

f(x) = 0

-(x-1)(x-5) = 0

x-1 = 0 or x-5 = 0

x = 1 or x = 5

The roots average to (x1+x2)/2 = (1+5)/2 = 3

The axis of symmetry is x = 3. This is enough to see that the vertex cannot possibly be (2,-9).

The axis of symmetry always passes through the vertex. It is the vertical line that cuts the parabola into two mirrored halves.

Put another way, the x coordinate of the vertex and the axis of symmetry are the same value.

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step-by-step explanation:

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