Answer
A) Vertex :$(0,8)$
B) Vertical intercept: $(0,8)$
Horizontal intercept : $\left(-\frac{2\sqrt 6}{3} ,0\right),\left(\frac{2 \sqrt{6}}{3}, 0\right)$
C) See the graph
D) Domain: All real numbers.
Range: $(-\infty, 8] $
Work Step by Step
Given\begin{equation}
\begin{aligned}
b(w)&=-3 w^2+8\\
a& = -3,\quad b= 0, \quad c= 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 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}
w & =\frac{-b}{2 a} \\
& =\frac{-(0)}{2(-3)} \\
& =0\\
f(0) & =c \\
& =8
\end{aligned}
\end{equation} The vertex of the function is $(0,8)$.
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& = 8.
\end{aligned}
\end{equation}Vertical intercept: $(0,8)$.
2) Find the horizontal intercept by setting the function to zero and solve.
\begin{equation}
\begin{aligned}
-3 w^2+8 & =0 \\
-3 w^2 & =-8 \\
w^2 & =\frac{8}{3} \\
w & = \pm \sqrt{\frac{8}{3}} \\
& = \pm \sqrt{\frac{4 \cdot 2}{3}} \\
& = \pm 2 \frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} \\
& = \pm \frac{2 \sqrt{6}}{3}.
\end{aligned}
\end{equation} Horizontal intercepts: $\left(-\frac{2\sqrt 6}{3} ,0\right),\left(\frac{2 \sqrt{6}}{3}, 0\right)$
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, 8] $
