How do you find the vertex of $f (x) = 8 - {(x + 2)}^{2}$ $f(x) = 8 - (x + 2) ^2$ ?

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How do you find the vertex of $f (x) = 8 - {(x + 2)}^{2}$ $f(x) = 8 - (x + 2) ^2$ ?

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Answer 1 · 2018-05-01T21:53:48

Vertex at $(- 2, 8)$ $(-2,8)$

Explanation:

Note the standard vertex form of a parabola is
$XXX f (x) = m {(x - a)}^{2} + b$ $color(white)("XXX")f(x)=color(green)m(x-color(red)a)^2+color(blue)b$ with vertex at $(a, b)$ $(color(red)a,color(blue)b)$

The given function: $f (x) = 8 - {(x + 2)}^{2}$ $f(x)=8-(x+2)^2$
an easily be re-arranged into standard vertex form as
$XXX f (x) = (- 1) {(x - (- 2))}^{2} + 8$ $color(white)("XXX")f(x)=color(green)(""(-1))(x-color(red)(""(-2)))^2+color(blue)8$ with vertex at $(- 2, 8)$ $(color(red)(-2),color(blue)8)$

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How do you find the vertex of $f (x) = 8 - {(x + 2)}^{2}$ $f(x) = 8 - (x + 2) ^2$ ?

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How do you find the vertex of $f (x) = 8 - {(x + 2)}^{2}$ $f(x) = 8 - (x + 2) ^2$ ?