# How do you find the vertex and intercepts for y = -2(x+1)^2 +7 ?

Jan 14, 2016

Explanation is given below.

#### Explanation:

The given problem is already in the vertex form.

$\textcolor{b l u e}{\text{The vertex form}}$

$\textcolor{m a r \infty n}{y = a {\left(x - h\right)}^{2} + k}$

Where $\left(h , k\right)$ is the vertex.

Our problem
$y = - 2 {\left(x + 1\right)}^{2} + 7$
$y = - 2 {\left(x - \left(- 1\right)\right)}^{2} + 7$

$\left(h , k\right) = \left(- 1 , 7\right)$

The vertex is $\left(- 1 , 7\right)$

Intercepts on $x$ and $y$ axes occur where the curve crosses them.

To find $y$ intercept we need to plug in $x = 0$

$y = - 2 {\left(0 + 1\right)}^{2} + 7$
$y = - 2 \left(1\right) + 7$
$y = - 2 + 7$
$y = 5$

The $y -$intercept is $\left(0 , 5\right)$

For finding $x -$intercepts, we need to plug in $y = 0$

$0 = - 2 {\left(x + 1\right)}^{2} + 7$
Subtract $7$ from both ends and isolating the term containing $x$
$- 7 = - 2 {\left(x + 1\right)}^{2}$

Let us rewrite it as $- 2 {\left(x + 1\right)}^{2} = - 7$ It looks better to when the variable is kept of the left side of the equation.

$- 2 {\left(x + 1\right)}^{2} = - 7$ dividing by $- 2$ on both sides isolates ${\left(x + 1\right)}^{2}$

We get

${\left(x + 1\right)}^{2} = \frac{7}{2}$

Take square root on both the sides we get

$\sqrt{{\left(x + 1\right)}^{2}} = \pm \sqrt{\frac{7}{2}}$
$x + 1 = \pm \sqrt{\frac{7}{2}}$

Subtract $1$ from both sides to solve for $x$

$x = - 1 \pm \sqrt{\frac{7}{2}}$

The $x -$intercepts are $\left(- 1 + \sqrt{\frac{7}{2}}\right)$ and $\left(- 1 - \sqrt{\frac{7}{2}}\right)$