Answer
A) Vertex: $(-2 ,17)$
B) Vertical intercept: $(0,15.8)$
Horizontal intercept: $(-9.5277,0),(5.5277,0)$
C) See the graph
D) Domain: All real numbers.
Range:$(-\infty, 17] $
Work Step by Step
Given \begin{equation}
W(g)=-0.3(g+2)^2+17.
\end{equation} 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 down since $a$ is negative.
Part A)
Step 2: Find the vertex. This function is in vertex form for a quadratic, so the vertex is
Part B)
Step 3: Find the vertical and horizontal intercepts.
1) Find the vertical intercept: \begin{equation}
\begin{aligned}
y& =-0.3(0+2)^2+17=15.8
\end{aligned}
\end{equation} Vertical intercept: $(0,15.8)$.
2) Find the horizontal intercept by setting the function to zero and solve. \begin{equation}
\begin{aligned}
-0.3(g+2)^2+17& =0 \\
(g+2)^2+17& =\frac{-17 }{-0.3}\\
g+2& = \pm \sqrt{\frac{170}{3}}\\
g& = -2\pm \sqrt{\frac{170}{3}}\\
\end{aligned}
\end{equation} Find the solutions: \begin{equation}
\begin{aligned}
& g=-2- \sqrt{\frac{170}{3}} \approx-9.5277 \\
& g=-2+ \sqrt{\frac{170}{3}} \approx 5.5277
\end{aligned}
\end{equation} Horizontal intercepts: $(-9.5277,0),(5.5277,0)$.
Part C) Sketch the graph as shown in the figure.
Part D) The domain and range of the function are given below:
Domain: All real numbers.
Range:$(-\infty, 17] $
