How do you find the x and y intercepts for $f (x) = {(x - 3)}^{2} + 17$ $f(x) = (x-3)^2 + 17$ ?

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How do you find the x and y intercepts for $f (x) = {(x - 3)}^{2} + 17$ $f(x) = (x-3)^2 + 17$ ?

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Answer 1 · 2017-08-02T13:29:43

$y-intercept = 26 no x-intercepts$ $"y-intercept "=26" no x-intercepts"$

Explanation:

$to find the intercepts$ $"to find the intercepts"$

$∙ let x = 0, in the equation for y-intercept$ $• " let x = 0, in the equation for y-intercept"$

$∙ let y = 0, in the equation for x-intercepts$ $• " let y = 0, in the equation for x-intercepts"$

$x = 0 \to y = {(- 3)}^{2} + 17 = 26 \leftarrow y-intercept$ $x=0toy=(-3)^2+17=26larrcolor(red)" y-intercept"$

$y = 0 \to {(x - 3)}^{2} + 17 = 0$ $y=0to(x-3)^2+17=0$

$\Rightarrow {(x - 3)}^{2} = - 17$ $rArr(x-3)^2=-17$

$AAx inRRcolor(white)(x) (x-3)^2>=0$

$rArr" there are no x-intercepts"$
graph{(x-3)^2+17 [-80, 80, -40, 40]}

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How do you find the x and y intercepts for $f (x) = {(x - 3)}^{2} + 17$ $f(x) = (x-3)^2 + 17$ ?

1 Answer

Explanation:

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How do you find the x and y intercepts for f(x) = (x-3)^2 + 17f(x)=(x−3)2+17f(x) = (x-3)^2 + 17?

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Explanation:

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How do you find the x and y intercepts for $f (x) = {(x - 3)}^{2} + 17$ $f(x) = (x-3)^2 + 17$ ?