What are he intercepts of $y = {(x + 1)}^{2} - 2$ $y=(x+1)^2-2$ ?

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What are he intercepts of $y = {(x + 1)}^{2} - 2$ $y=(x+1)^2-2$ ?

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Answer 1 · 2018-07-25T04:24:56

The $x$ $x$ -intercepts are at $(\sqrt{2} - 1)$ $(sqrt2-1)$ and $(- \sqrt{2} - 1)$ $(-sqrt2-1)$ and the $y$ $y$ -intercept is at $(0, - 1)$ $(0, -1)$ .

Explanation:

To find the $x$ $x$ -intercept(s), plug in $0$ $0$ for $y$ $y$ and solve for $x$ $x$ .

$0 = {(x + 1)}^{2} - 2$ $0 = (x+1)^2 - 2$

Add $2$ $color(blue)2$ to both sides:
$2 = {(x + 1)}^{2}$ $2 = (x+1)^2$

Square root both sides:
$\pm \sqrt{2} = x + 1$ $+-sqrt2 = x+1$

Subtract $1$ $color(blue)1$ from both sides:
$\pm \sqrt{2} - 1 = x$ $+-sqrt2 - 1 = x$

Therefore, the $x$ $x$ -intercepts are at $(\sqrt{2} - 1)$ $(sqrt2-1)$ and $(- \sqrt{2} - 1)$ $(-sqrt2-1)$ .

To find the $y -$ $y-$ intercept, plug in $0$ $0$ for $x$ $x$ and solve for $y$ $y$ :
$y = {(0 + 1)}^{2} - 2$ $y = (0+1)^2 - 2$

Simplify:
$y = 1^{2} - 2$ $y = 1^2 - 2$

$y = 1 - 2$ $y = 1 - 2$

$y = - 1$ $y = -1$

Therefore, the $y$ $y$ -intercept is at $(0, - 1)$ $(0, -1)$ .

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