Answer
A) Vertex : $(4.5 ,-52.9)$
B) Vertical intercept: $(0,-44.8)$
Horizontal intercepts: $(-7,0),(16,0)$
C) See the graph
D) Domain: All real numbers.
Range: $[-52.9,\infty )$
Work Step by Step
Given
\begin{equation}
\begin{aligned}
h(w)&=0.4 w^2-3.6 w-44.8\\
a& = 0.4 ,\quad b= -3.6, \quad c= -44.8.
\end{aligned}
\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 up since $a$ is positive.
Part A)
Step 2: Find the vertex. This function is in standard form for a quadratic, so we can use the formula: \begin{equation}
\begin{aligned}
& w=\frac{-b}{2 a}=\frac{-(-3.6)}{2(0.4)}=4.5 \\
& \begin{aligned}
w(4.5) & =0.4 \cdot 4 .5^2-3.6\cdot 4 .5-44.8 \\
& =-52.9.
\end{aligned}
\end{aligned}
\end{equation} The vertex of the function is $(4.5 ,-52.9)$.
Part B)
Step 3: Find the vertical and horizontal intercepts.
1) Find the vertical intercept which is equal to the constant $c$ if the parabola is in its standard form. \begin{equation}
\begin{aligned}
y& = -44.8.
\end{aligned}
\end{equation}
Vertical intercept: $(0,-44.8)$.
2) Find the horizontal intercept by setting the function to zero and solve. \begin{equation}
\begin{aligned}
& 0.4 w^2-3.6 w-44.8=0 \\
& \left(0.4 w^2-3.6 w-44.8\right) \cdot 10=0.10 \\
& 4 w^2-36 w-448=0
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
&w=\frac{-(-36) \pm \sqrt{36^2-4(4)(-448)}}{2(4)} \\
&w=\frac{+36 \pm \sqrt{8464}}{8}\\
&=\frac{36 \pm 92}{8}\\
&=\frac{36 \pm 92}{8}
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
& w=\frac{36-92}{8}=-7 \\
& w=\frac{36+92}{8}=16.
\end{aligned}
\end{equation} Horizontal intercepts: $(-7,0),(16,0)$.
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: $[-52.9,\infty )$
