Question

In the following exercises $y=-2x^2-4x-5$,

Part (a) write the equation in standard form.

Part (b) use properties of the standard form to graph the equation.

Short answer

Part (a): The required equation is $y=-2{\left(x+1\right)}^2-3$.

Part (b): The required graph is given below,

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given equation,

$y=-2x^2-4x-5$

Rewrite the function in $f\left(x\right)=a{\left(x-h\right)}^2+k$form by completing the square,

$$\begin{gathered} y=-2\left(x^2+2x\right)-5 \\ y=-2\left(x^2+2x+1\right)-5+2 \\ y=-2{\left(x+1\right)}^2-3......\left(i\right) \end{gathered}$$

Part (b) Step 1. Graph the function using transformation.

Identify the constants,

$$\begin{gathered} a=-2, \\ h=-1, \\ k=-3 \end{gathered}$$

Since $a=-2$, the parabola opens downward.

The axis of symmetry is $x=h$. Then,

$x=-1$.

The vertex is localid="1645204103192" $\left(h,k\right)$. Then $\left(-1,-3\right)$.

Substitute $x=0$in equation (i),

$$\begin{gathered} y=-2{\left(0+1\right)}^2-3 \\ y=-2-3 \\ y=-5 \end{gathered}$$

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

Part (b) Step 2. Find the symmetric point.

The point symmetric to $\left(0,-5\right)$across the axis of symmetry is \(x= -1\).

Substitute $y=0$in equation (i),

$$\begin{gathered} 0=-2{\left(x+1\right)}^2-3 \\ 0=-2x^2-4x-5 \end{gathered}$$

The discriminant negative, so there are nox-intercepts.

Plot the graph,