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

Apr 1, 2017

In vertex form, the vertex is given as the $h$ and $k$ values.

The intercepts, just sub in each variable as $0$ and solve for the other.

#### Explanation:

The equation is given to us in vertex form, making the process of determining the vertex much easier.

The base formula for vertex form is $f \left(x\right) = a \left(x - h\right) + k$
=> Where $h$ is the x-coordinate of the vertex.
=> Where $k$ is the y-coordinate of the vertex.
=> Where $a$ is factor of how much the parabola stretches/compresses by in a vertical sense.

The "$\left(x + 2\right)$" gives us the x-coordinate of the vertex as $x = - 2$. There is no $k$-value, meaning the y-coordinate is on the x-axis.

Therefore, the vertex of the equation is $\left(- 2 , 0\right)$.

The intercepts can be determined by simply subbing in each of the variables as 0, and solving for the other.

First, we'll find the x-intercept. Therefore, we'll sub in the y-value as $0$.

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

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

I switched $f \left(x\right)$ into $y$ to make comprehension easier

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

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

$0 = x + 2$

$- 2 = x$

And now the y-intercept. Therefore, we'll sub in the x-value as $0$.

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

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

I switched $f \left(x\right)$ into $y$ to make comprehension easier

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

$= - 2 {\left(2\right)}^{2}$

$= - 2 \left(4\right)$

$= - 8$

Your parabola will look like this.

$f \left(x\right) = - 2 {\left(x + 2\right)}^{2}$ graph{-2(x+2)^2 [-10.875, 9.125, -6.28, 3.72]}

As you can see, the vertex is in fact $\left(- 2 , 0\right)$, and the intercepts are $\left(- 2 , 0\right)$ and $\left(0 , - 8\right)$.

Hope this helps :)