How do you find the x-intercepts of a quadratic function in standard form given $f (x) = 2 {(x + 1)}^{2} - 3$ $f(x) = 2(x+1)^2 - 3$ ?

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How do you find the x-intercepts of a quadratic function in standard form given $f (x) = 2 {(x + 1)}^{2} - 3$ $f(x) = 2(x+1)^2 - 3$ ?

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Answer 1 · 2018-03-29T05:07:31

$x = - 1 \pm \frac{\sqrt{6}}{2}$ $x = - 1 +- sqrt6/2$

Explanation:

This function is written in vertex form.
To find the x-intercepts, make f(x) = 0.
$2 {(x + 1)}^{2} - 3 = 0$ $2(x + 1)^2 - 3 = 0$
${(x + 1)}^{2} = \frac{3}{2}$ $(x + 1)^2 = 3/2$
$x + 1 = \pm \frac{\sqrt{3}}{\sqrt{2}}$ $x + 1 = +- sqrt3/sqrt2$
$x = - 1 \pm \frac{\sqrt{3}}{\sqrt{2}}$ $x = - 1 +- sqrt3/sqrt2$ , or
$x = - 1 \pm \frac{\sqrt{6}}{2}$ $x = - 1 +- sqrt6/2$

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How do you find the x-intercepts of a quadratic function in standard form given $f (x) = 2 {(x + 1)}^{2} - 3$ $f(x) = 2(x+1)^2 - 3$ ?

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How do you find the x-intercepts of a quadratic function in standard form given f(x) = 2(x+1)^2 - 3f(x)=2(x+1)2−3f(x) = 2(x+1)^2 - 3?

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Explanation:

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How do you find the x-intercepts of a quadratic function in standard form given $f (x) = 2 {(x + 1)}^{2} - 3$ $f(x) = 2(x+1)^2 - 3$ ?