Question

In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.

$y=-2x^2-8x-12$

Short answer

The given information is:

$y=-2x^2-8x-12$

Step-by-step solution

Step 1. Given information.

The given information is:

$y=-2x^2-8x-12$

Step 2. Find the axis of symmetry and the vertex.

$$\begin{gathered} y=a{x}^2+bx+c \\ y=-2x^2-8x-12 \end{gathered}$$

The value of coefficient ais negative therefore we can say that the parabola opens down.

The axis of symmetry is the line$x=-\frac{b}{2a}$.

Substitute the values of a, and binto the equation.

$$\begin{gathered} x=-\frac{\left(-8\right)}{2\left(-2\right)} \\ =-2 \end{gathered}$$

Therefore, the axis of symmetry is $x=-2$.

The vertex is on the line of symmetry, so its x-coordinate will be $x=-2$.

Now substitute the value of xinto the equation,

$$\begin{gathered} y=2{\left(-2\right)}^2-8\left(-2\right)-12 \\ =2\left(4\right)+16-12 \\ =-8+4 \\ =-4 \end{gathered}$$

Therefore, the vertex is$\left(-2,-4\right)$.

Step 3. Find the intercepts.

Now substitute xequal to 0 in the equation to find the intercept y,

$$\begin{gathered} y=-2{\left(0\right)}^2-8\left(0\right)-12 \\ =-0-0-1 \\ =-12 \end{gathered}$$

Therefore, the point $\left(0,-12\right)$ is the y-intercept.

We can see that the resultant point is right of the line of symmetry by 2 units.

Therefore the point right to the line of symmetry by 2 units is $\left(-4,-12\right)$.

Now substitute yequal to 0 in the equation to find the intercept x,

$$\begin{gathered} -2x^2-8x-12=0 \\ x=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)} \\ x=\frac{8\pm \sqrt{-36}}{-4} \end{gathered}$$

There is no x-intercept because a square root is not possible for a negative number.

Step 4. Plot the graph.