Answer
Part A) Vertex: $(0,1)$
Part B) Vertical intercept: $(0, 1)$
Horizontal intercept: None
Part C) See graph
Part D) Domain: All real numbers.
Range:$[1,\infty )$
Work Step by Step
Given\begin{equation}
\begin{aligned}
g(n)&=\frac{2}{3} n^2+1\\
a& = \frac{2}{3} ,\quad b= 0, \quad c= 1.
\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}
n & =\frac{-b}{2 a} \\
& =\frac{-(0)}{2\cdot 2/3} \\
& =0\\
f(0) & =c \\
& =1.
\end{aligned}
\end{equation} The vertex of the function is $(0,1)$.
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}
g& = 1
\end{aligned}
\end{equation} Vertical intercept: $(0,1)$.
2) Find the horizontal intercept by setting the function to zero and solve.
\begin{equation}
\begin{aligned}
\frac{2}{3} n^2+1 & =0 \\
\frac{2}{3} n^2 & =-1 \\
n^2 & =-\frac{3}{2}.
\end{aligned}
\end{equation} This equation has no solution. Hence, there is no horizontal intercept.
Part C) See graph.
Part D) The domain and range are as follows:
Domain: All real numbers
Range: $[1,\infty)$.
