Question

Find the vertex of this quadratic function:
r(x)=-4x^2+16x-15

Answer 1

@satellite73

Answer 2

@Zarkon

Answer 3

first coordinate of the vertex is \(-\frac{b}{2a}\)

Answer 4

in your case that is
\[-\frac{16}{2\times (-4)}=2\]

Answer 5

second coordinate of the vertex is what you get when you replace \(x\) by the first coordinate

Answer 6

no i dont get how to set up the equation like what i did was add 15 to both sides

Answer 7

@ganeshie8

Answer 8

\[r(x) =-4x^2 + 16x - 15\]
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Use vertex form
\[y = a(x-h)^2 + k\]
where (h,k) are the coordinates of the vertex
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Group thru parenthesis
\[r(x) = (-4x^2 + 16x) - 15\]
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factor out -4
\[r(x) = -4(x^2 + 4x) - 15\]
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Complete the square by adding and subtracting 16
\[r(x) = -4(x^2 + 4x + 4) - 15 + 16\]
Notice that we only add 4 in the parenthesis since when -4 is distributed, it will become -16
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\[r(x) = -4(x^2 + 4x + 4) - 15 + 16\]
\[r(x) = -4(x+2) + 1\]
h=-2, k=1
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Therefore the coordinates of the vertex is (-2,1).

Answer 9

* \[r(x)=-4(x+2)^2+1\]