Question #c2ac4

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Question #c2ac4

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Answer 1 · 2017-12-01T00:46:22

The vertex form is $f (x) = {(x - \frac{9}{2})}^{2} - \frac{81}{4}$ $f(x) = (x-9/2)^2-81/4$ .

Explanation:

Vertex form is $f (x) = a {(x - h)}^{2} + k$ $f(x) = a(x-h)^2+k$

The vertex $(h, k)$ $(h,k)$ of a quadratic $a x^{2} + b x + c$ $ax^2+bx+c$ can be found from $h = - \frac{b}{2 a}$ $h=-b/(2a)$ and $k = f (h)$ $k=f(h)$ .

For this quadratic $a = 1, b = - 9$ $a=1,b=-9$ , and $c = 0$ $c=0$ . So we know that $h = - \frac{- 9}{2 (1)} = \frac{9}{2}$ $h=-(-9)/(2(1))=9/2$ .

$k = f (\frac{9}{2}) = {(\frac{9}{2})}^{2} - 9 (\frac{9}{2}) = \frac{81}{4} - \frac{81}{2} = - \frac{81}{4}$ $k=f(9/2) = (9/2)^2-9(9/2) = 81/4-81/2 = -81/4$ .

The vertex form is $f (x) = {(x - \frac{9}{2})}^{2} - \frac{81}{4}$ $f(x) = (x-9/2)^2-81/4$ .

Answer 2 · 2017-12-01T00:51:27

f(x)= ${(x - \frac{9}{2})}^{2}$ $(x-9/2)^2$ - $\frac{81}{4}$ $81/4$

Explanation:

The vertex is ( h,k )

h can be found by $- \frac{b}{2 a}$ $-b/(2a)$

After finding h you can insert it into the original equation

After doing so you will get - $\frac{81}{4}$ $81/4$ as k.

Now that you have the vertex ( $\frac{9}{2}$ $9/2$ , - $\frac{81}{4}$ $81/4$ ) pug them into Vertex form: F(x) = ${(x - h)}^{2}$ $(x-h)^2$ +k

This can also be solved by completing the square.

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