Answer
1) Vertex: $(-9,-15)$.
2) Vertical intercept: $(0,147)$
Horizontal intercepts: $(-11.7386 ,0),(-6.2614,0)$
3) Domain: All real numbers.
Range: $[-15, \infty) $
Work Step by Step
Given \begin{equation}
f(x)=2(x+9)^2-15.
\end{equation}
The function is in the vertex form $$f(x)=a(x-h)^2+k,$$ where $$a=2,h=-9,k=-15.$$ Step 1: Determine whether the graph opens up or down. The parabola opens up when the constant, $a$, is positive and opens down, when $a$ negative. This parabola opens up since $a$ is positive.
The vertex of the function is $(-9 ,-15)$ since the parabola is in standard vertex form.
Step 3: Find the vertical and horizontal intercepts.
1) Find the vertical intercept which is found by setting $x= 0$: \begin{equation}
\begin{aligned}
y& =2(0+9)^2-15=147
\end{aligned}
\end{equation} Vertical intercept: $(0,147)$.
2) Find the horizontal intercept by setting the function to zero and solve. \begin{equation}
\begin{aligned}
2(x+9)^2-15& =0 \\
((x+9)^2& =\frac{15 }{2}\\
x+9& = \pm \sqrt{7.5}\\
x& = -9\pm \sqrt{6}\\
& x=-9- \sqrt{7.5}\approx-11.7386 \\
& x=-9+ \sqrt{7.5}\approx -6.2614.
\end{aligned}
\end{equation} The graph of the function is shown in the figure with the required points shown.
