Question

Y=x² + 16x + 64

Answer 1

I'll do the first one to get you started

The equation y = x^2+16x+64 is the same as y = 1x^2+16x+64

Compare that to y = ax^2+bx+c and we see that

a = 1

b = 16

c = 64

Use the values of 'a' and b to get the value of h as shown below

h = -b/(2a)

h = -16/(2*1)

h = -8

This is the x coordinate of the vertex.

Plug this x value into the original equation to find the corresponding y value of the vertex.

y = x^2+16x+64

y = (-8)^2 + 16(-8) + 64

y = 0

Since the y coordinate of the vertex is 0, this means k = 0.

The vertex is (h,k) = (-8, 0)

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So we found that a = 1, h = -8 and k = 0

Therefore,

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

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

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

is the vertex form

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Final answer to problem 1 is f(x) = (x+8)^2