# How do you graph the quadratic function and identify the vertex and axis of symmetry for y=-2(x+3)^2-4?

Sep 4, 2016

See below.

#### Explanation:

This quadratic function will have a vertex at $\left(- 3 , - 4\right)$, since in vertex form $y = a {\left(x - p\right)}^{2} + q$, the vertex is located at $\left(p , q\right)$.

The equation of the axis of symmetry (it is a vertical line) will be $x = - 3$. That will always be the x-coordinate of the vertex.

To graph, it is not only helpful to know the vertex and the equation of the axis of symmetry. Intercepts, both x and y, and direction of opening is extremely important.

Let's start with the latter. The parameter $a$ in $y = a {\left(x - p\right)}^{2} + q$ influences the breadth and the direction of opening of the parabola. If $a > 0$, then the parabola opens upwards. Similarly, if $a < 0$, the parabola opens down.

Here, $a = - 2$, so the parabola opens down.

Now for x-intercepts. These can be obtained by setting $y = 0$ and solving the resulting quadratic.

$y = - 2 {\left(x + 3\right)}^{2} - 4$

$0 = - 2 {\left(x + 3\right)}^{2} - 4$

$\frac{4}{-} 2 = {\left(x + 3\right)}^{2}$

$\pm \sqrt{- 2} = x + 3$

$x = \emptyset$

So, there is no x-intercept. We could have figured out without algebra, because the vertex lies below the x-axis and the parabola opens downwards.

As for y-intercepts, set $x = 0$ and solve.

$y = - 2 {\left(x + 3\right)}^{2} - 4$

$y = - 2 {\left(0 + 3\right)}^{2} - 4$

$y = - 2 \left(9\right) - 4$

$y = - 22$

$\therefore$ The y-intercept has coordinates of $\left(0 , - 22\right)$.

Let's finish by identifying the domain and range. The domain is $x \in \mathbb{R}$ and the range is y ≤ -4, since the point $\left(- 3 , - 4\right)$ is the maximum.

We can now graph, using a table of values and connecting the points using a smooth, curved line.

Here is what your graph should look like:

Hopefully this helps!