How do you rewrite into vertex form #f(x)=4x^2-16x+5#?

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Answer 1 · 2016-11-14T23:23:56

Please see the explanation.

The vertex form for a parabola that opens or down is:

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

Please observe that, in the given equation, #a = 4#, therefore we add zero to the original equation in the form of #4h^2 - 4h^2#:

#f(x) = 4x^2 - 16x + 4h^2 - 4h^2 + 5#

Factor 4 out of the first 3 terms and group them with ()s:

#f(x) = 4(x^2 - 4x + h^2) - 4h^2 + 5#

We can find the value of h by setting the middle term equal to -2hx:

#-2hx = -4x#

#h = 2# and #-4h^2 = -16#

#f(x) = 4(x^2 - 4x + h^2) - 16 + 5#

We know that three terms inside the ()s is equal to #(x - 2)^2#

#f(x) = 4(x - 2)^2 - 11#