How do you complete the square of f(t) = -16t^2 - 64t + 80 to find the vertex?

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I know that the vertex of the quadratic equation is (-2,144).

My work:

f(t) = -16t^2 - 64t + 80

f(t) = -16(t^2 + 4t) + 80 (I divided the x terms by the leading coefficient and grouped them from the constant.)

f(t) = -16(t^2 + 4t + 2^2) + 80 - 4 (I divided the middle term's coefficient by two and squared it, adding it inside the parentheses and subtracting it outside.)

f(t) = -16(t - 2)^2 + 76 (This is the part I most likely messed up on.)

Jul 31st, 2015

f(t) = -16t^2 - 64t + 80

We can write it as 

y = - 16t^2 - 64t + 80

Now

y - 80 = - 16t^2 - 64t

y - 80 = - 16(t^2 + 4t)

Now 

y - 80 - 16___ = - 16(t^2 + 4t + ___)

WE put same number on both side blanks, then

y - 80 - 16(4) = - 16(t^2 + 4t + 4)

y - 80 - 64 = - 16(t + 2)^2

y - 144 = - 16(t + 2)^2

y = - 16(t + 2)^2 + 144

So the exact form of the vertex form is 

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

So by comparing above with it, we get

y = - 16(t - (-2))^2 + 144

So the vertex is

(-2, 144)

Jul 31st, 2015

I see what I did wrong. Thanks a lot!

Jul 31st, 2015

My pleasure. : )

Jul 31st, 2015

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Jul 31st, 2015
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