Answer
Vertex: $(-10,15)$
Vertical intercept: $(0,-185)$
Horizontal intercepts: $(-12.74,0),(-7.26,0)$
Work Step by Step
Given $$
f(x)=-2(x+10)^2+15.
$$ The parabola is in its standard vertex form, $f(x) = a(x-h)^2+k$. The vertex can be directly read from the equation. The vertex is $(-10,15)$. The horizontal intercept is found by setting the $y$ component of the parabola to zero and solving for $x$.
$$
\begin{aligned}
-2(x+10)^2+15&=0 \\
-2(x+10)^2&=-15 \\
\frac{-2(x+10)^2}{-2}&=\frac{15}{-2} \\
(x+10)^2& = 7.5\\
x+10&= \pm\sqrt{7.5}\\
x &=-10 \pm\sqrt{7.5}\\
\textbf{This gives:}\\
x&= -10 -\sqrt{7.5}\\
&\approx -12.74\\
x&= -10 +\sqrt{7.5}\\
&\approx -7.26.
\end{aligned}
$$ The horizontal intercepts are $(-12.74,0)$ and $(-7.26,0)$.
The graph has a vertical intercept for $x= 0$ Find the value of $y$ at $x= 0$.
$$ \begin{aligned}
f(0)&=-2(0+10)^2+15 \\
y&=-185.
\end{aligned}
$$ The vertical intercept is $(0,-185)$.
