Answer
A) Vertex: $(-4 ,86.8)$
B) Vertical intercept: $(0,82)$
Horizontal intercepts: $(-21,0),(13,0)$
C) See the graph
D) Domain: All real numbers.
Range: $(-\infty, 86.8] $
Work Step by Step
Given \begin{equation}
\begin{aligned}
Q(p)&=-0.3 p^2-2.4 p+82\\
a & =-0.3, \quad b=-2.4, \quad c=82
\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 down since $a$ is negative.
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}
&p=\frac{-b}{2 a}=-\frac{-2.4}{2\cdot(-0.3)}=-4 \\
& \begin{aligned}
Q(-4) & =-0.3(-4)^2-2.4(-4)+82 \\
& =86.8
\end{aligned}
\end{aligned}
\end{equation} The vertex of the function is $(-4 ,86.8)$.
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& =82.
\end{aligned}
\end{equation} Vertical intercept: $(0,82)$.
2) Find the horizontal intercept by setting the function to zero and solve.
\begin{equation}
\begin{aligned}
-0.3 p^2-2.4 p+82& =0
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
p & =\frac{-(-2.4) \pm \sqrt{(-2.4)^2-4(-0.3)(82)}}{2 \cdot(-0.3)} \\
& =-\frac{2.4 \pm \sqrt{104.16}}{0.6} \\
&\approx -( 4 \pm 17)
\end{aligned}
\end{equation} The solutions are: \begin{equation}
\begin{aligned}
& p=-( 4 + 17)=-21 \\
& p=-( 4 - 17)=13.
\end{aligned}
\end{equation} Horizontal intercepts: $(-21,0),(13,0)$.
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, 86.8] $
