Answer
A) Vertex :$(0,5)$
B) Vertical intercept: $(0,5)$
Horizontal intercept: None
C) See the graph
D) Domain: All real numbers.
Range: $[5,\infty )$
Work Step by Step
Given \begin{equation}
\begin{aligned}
f(x)&=2 x^2+5\\
a& = 2,\quad b= 0,\quad c= 5.
\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}
x & =\frac{-b}{2 a} \\
& =\frac{-(0)}{2(4)} \\
& =0\\
f(0) & =c \\
& =5.
\end{aligned}
\end{equation}The vertex of the function is $(0,5)$.
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}
h& = 5
\end{aligned}
\end{equation} Vertical intercept: $(0,5)$.
2) Find the horizontal intercept by setting the function to zero and solve.
\begin{equation}
\begin{aligned}
2 x^2+5 & =0 \\
2 x^2 & =-5 \\
x^2 & =-\frac{5}{2}
\end{aligned}
\end{equation} There is no horizontal intercept for this function.
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:$[5,\infty )$
