Answer
A) Vertex: $(0,12)$
B) Vertical intercept: $(0,12)$
Horizontal intercept: $(-2 \sqrt{6}, 0),(2 \sqrt{6}, 0)$
C) See the graph
D) Domain: All real numbers.
Range: $(-\infty, 12] $
Work Step by Step
Given \begin{equation}
\begin{aligned}
s(x)&=-\frac{1}{2} x^2+12\\
a& = -\frac{1}{2} ,\quad b= 0, \quad c= 12.
\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}
x & =\frac{-b}{2 a} \\
& =\frac{-(0)}{2\cdot (-1/2)} \\
& =0\\
h(0) & =c \\
& =12.
\end{aligned}
\end{equation} The vertex of the function is $(0,12)$.
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}
s& = 12.
\end{aligned}
\end{equation} Vertical intercept: $(0,12)$.
2) Find the horizontal intercept by setting the function to zero and solve.
\begin{equation}
\begin{aligned}
\frac{-1}{2} x^2+12 & =0 \\
-\frac{1}{2} x^2 & =-12 \\
x^2 & =24 \\
x & = \pm \sqrt{4 \cdot 6} \\
x & = \pm 2 \sqrt{6}.
\end{aligned}
\end{equation} Horizontal intercept $(-2 \sqrt{6}, 0),(2 \sqrt{6}, 0)$.
Horizontal intercept: None
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:$(-\infty ,12]$
