Answer
A) Vertex :$(-
15.5,-264.5)$
B) Vertical intercept: $(0,216)$
Horizontal intercepts: $(4,0), (27,0)$
C) See the graph
D) Domain: All real numbers.
Range: $[-264.5,\infty )$
Work Step by Step
Given\begin{equation}
\begin{aligned}
g(s)&=2 s^2-62 s+216\\
a& = 2,\quad b= -62, \quad c= 216.
\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}
s & =\frac{-b}{2 a} \\
& =\frac{-(-62)}{2(2)} \\
& =15.5\\
g(15.5) & =2(15.5)^2-62(15.5)+216 \\
& =-264.5
\end{aligned}
\end{equation} The vertex of the function is $(15.5,-264.5)$.
Part B)
Step 3: Find the vertical and horizontal intercepts.
1) Find the vertical intercept which equal to the constant $c$ if the parabola is in its standard form.
\begin{equation}
\begin{aligned}
s& = 216
\end{aligned}
\end{equation}Vertical intercept: $(0,216)$.
2) Find the horizontal intercept by setting the function to zero and solving for $s$.
\begin{equation}
\begin{aligned}
&2s^2-62s+216=0\\
& s=\frac{-(-62)\pm\sqrt{(-62)^2-4(2)(216)}}{2 \cdot(2)} \\
& =\frac{62 \pm \sqrt{2116}}{4} \\
& =\frac{62 \pm 46}{4} \\
& =15.5 \pm 11.5\\
& s=15-11.5=4 \\
& s=15.5+11.5=27
\end{aligned}
\end{equation} Horizontal intercepts: $(4,0), (27,0)$.
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:$[-264.5,\infty )$
