Question

graph $f(x)=-2(x + 3)^{2}-4$.

Explanation:

Step1: Identify the vertex form

The vertex - form of a parabola is $y=a(x - h)^2+k$, where $(h,k)$ is the vertex. For the function $f(x)=-2(x + 3)^2-4$, we have $a=-2$, $h=-3$, and $k = - 4$. So the vertex is $(-3,-4)$.

Step2: Determine the direction of opening

Since $a=-2<0$, the parabola opens downwards.

Step3: Find the y - intercept

Set $x = 0$:
\[

$$\begin{align*} f(0)&=-2(0 + 3)^2-4\\ &=-2\times9-4\\ &=-18-4\\ &=-22 \end{align*}$$

\]
So the y - intercept is $(0,-22)$.

Step4: Find the x - intercepts (if any)

Set $y = 0$:
\[

$$\begin{align*} 0&=-2(x + 3)^2-4\\ 2(x + 3)^2&=-4\\ (x + 3)^2&=-2 \end{align*}$$

\]
Since the square of a real number is non - negative, there are no real x - intercepts.

To graph the function, plot the vertex $(-3,-4)$, the y - intercept $(0,-22)$, and use the fact that the parabola opens downwards and is symmetric about the line $x=-3$.

Answer:

Graph a parabola with vertex at $(-3,-4)$ that opens downwards, has a y - intercept at $(0,-22)$ and no real x - intercepts.