I know that the vertex of the quadratic equation is (-2,144).
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.)
We can write it as
y = - 16t^2 - 64t + 80
y - 80 = - 16t^2 - 64t
y - 80 = - 16(t^2 + 4t)
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
I see what I did wrong. Thanks a lot!
My pleasure. : )
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