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How do you write $f (x) = x^{2} + 2 x - 8$ $f(x) = x^2+2x-8$ in vertex form?

1 Answer

Apr 8, 2016

vertex form: ${(x - (- 1))}^{2} + (- 9)$ $(x-color(red)(color(white)("")(-1)))^2+color(blue)(color(white)("")(-9))$ with vertex at $(- 1, - 9)$ $(color(red)(-1),color(blue)(-9))$

Explanation:

General vertex form:
$XXX y = m {(x - a)}^{2} + b$ $color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)$ with vertex at $(a, b)$ $(color(red)(a),color(blue)(b))$

$f (x) = x^{2} + 2 x - 8$ $f(x)=x^2+2x-8$

$\to f (x) = x^{2} + 2 x + 1 - 9$ $rarr f(x)=x^2+2x+1 -9$

$\to f (x) = {(x + 1)}^{2} - 9$ $rarr f(x)=(x+1)^2-9$

$\to f (x) = 1 {(x - (- 1))}^{2} + (- 9) XXX$ $rarr f(x)=color(green)(1)(x-color(red)(color(white)("")(-1)))^2+color(blue)(""(-9))color(white)("XXX")$ (in complete vertex form)

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