How do you write $y = -4x^2 + 8x + 60$ into vertex form?

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Answer 1 · 2015-05-10T16:28:35

$y = -4x^2+8x+60 = -4(x-1)^2 + 64$ .

As I mentioned on another answer:

In general, for $y = ax^2 + bx + c$ , the $x$ coordinate of the vertex is $-b/(2a)$ , and the vertex form is:

$y = a(x + b/(2a))^2 + (c - b^2/(4a))$

Answer 2 · 2015-05-10T21:31:50

Another way:
Vertex form:
$f(x) = a(x -b/(2a))^3 + f(-b/(2a))$

-b/2a = -8/-8 = 1

f(1) = -4 + 8 + 60 = 64

Vertex form: $y = -4.(x - 1)^2 + 64$

Check:
$Develop f(x) = -4(x - 1)^2 + 64 = -4(x^2 - 2x + 1) + 64 = -4x^2 + 8x - 4 + 64 = -4x^2 + 8x + 60$ Correct

How do you write y = -4x^2 + 8x + 60y = -4x^2 + 8x + 60 into vertex form?