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

Jul 25, 2018

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

#### Explanation:

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

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

Add $\textcolor{b l u e}{2}$ to both sides:
$2 = {\left(x + 1\right)}^{2}$

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

Subtract $\textcolor{b l u e}{1}$ from both sides:
$\pm \sqrt{2} - 1 = x$

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

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

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

$y = 1 - 2$

$y = - 1$

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

Hope this helps!