Which function in vertex form is equivalent to f(x) = x2 + 8 – 16x?
A. f(x) = (x – 8)2 – 56
B. f(x) = (x – 4)2 + 0
C. f(x) = (x + 8)2 – 72
D.f(x) = (x + 4)2 – 32

To make the perfect square we add and subtract the square of [tex]\frac{b}{2a}[/tex], where b is coefficient of x and a is the coefficient of [tex]x^2[/tex].

Since a=1 and b = -16, So we will add and subtract he square of -8.

[tex]f(x)=(x^2-16x+(-8)^2)+8-(-8)^2[/tex]

[tex]f(x)=(x^2-16x+(8)^2)+8-64[/tex]

Using [tex](a-b)^2=a^2-2ab+b^2[/tex]

[tex]f(x)=(x-8)^2)-56[/tex]

Therefore, the correct option is A.

Which function in vertex form is equivalent to f(x) = x2 + 8 – 16x? A. f(x) = (x – 8)2 – 56 B. f(x) = (x – 4)2 + 0 C. f(x) = (x + 8)2 – 72 D.f(x) = (x + 4)2 – 32

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Which function in vertex form is equivalent to f(x) = x2 + 8 – 16x?
A. f(x) = (x – 8)2 – 56
B. f(x) = (x – 4)2 + 0
C. f(x) = (x + 8)2 – 72
D.f(x) = (x + 4)2 – 32