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

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Answer 1 · 2018-07-25T04:24:56

The $x$ -intercepts are at $(sqrt2-1)$ and $(-sqrt2-1)$ and the $y$ -intercept is at $(0, -1)$ .

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

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

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

Square root both sides:
$+-sqrt2 = x+1$

Subtract $color(blue)1$ from both sides:
$+-sqrt2 - 1 = x$

Therefore, the $x$ -intercepts are at $(sqrt2-1)$ and $(-sqrt2-1)$ .

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

Simplify:
$y = 1^2 - 2$

$y = 1 - 2$

$y = -1$

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

Hope this helps!

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