Answer
A) Vertex :$(0,3)$
B) Vertical intercept: $(0,3)$
Horizontal intercepts: $\left(-\sqrt{2}, 0\right),\left(\sqrt{2},0\right)$
C) See the graph
D) Domain: All real numbers.
Range: $(-\infty, 3] $
Work Step by Step
Given \begin{equation}
\begin{aligned}
d(p)&=-1.5 p^2+3\\
a& = -1.5,\quad b= 0, \quad c= 3.
\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{-(0)}{2(-3)} \\
& =0\\
f(0) & =c \\
& =3.
\end{aligned}
\end{equation} The vertex of the function is $(0,3)$.
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}
p& = 3.
\end{aligned}
\end{equation} Vertical intercept: $(0,3)$.
2) Find the horizontal intercept by setting the function to zero and solve.
\begin{equation}
\begin{aligned}
-1.5 p^2+3 & =0 \\
-1.5 p^2 & =-3 \\
p^2 & =\frac{3}{1.5} \\
p^2 & =2 \\
p & = \pm \sqrt{2}.
\end{aligned}
\end{equation} Horizontal intercepts: $\left(-\sqrt{2}, 0\right),\left(\sqrt{2}, 0\right)$
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, 3] $
