Answer
A) Vertex: $(4 ,-18)$
B) Vertical intercept: $(0,30)$
Horizontal intercepts: $(1.5505 ,0),(6.4495,0)$
C) See the graph
D) Domain: All real numbers.
Range: $[-18,\infty)$
Work Step by Step
Given \begin{equation}
f(x)=3(x-4)^2-18.
\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)
The vertex of the function is $(4 ,-18)$ since the parabola is in standard vertex form.
Part B)
Step 3: Find the vertical and horizontal intercepts.
1) Find the vertical intercept which is found by setting $x= 0$. \begin{equation}
\begin{aligned}
y& =3(0-4)^2-18= 30.
\end{aligned}
\end{equation} Vertical intercept: $(0,30)$.
2) Find the horizontal intercept by setting the function to zero and solve. \begin{equation}
\begin{aligned}
3(x-4)^2-18& =0 \\
(x-4)^2& =\frac{18 }{3}\\
x-4& = \pm \sqrt{6}\\
x& = 4\pm \sqrt{6}.
\end{aligned}
\end{equation} The solutions are: \begin{equation}
\begin{aligned}
& x=4- \sqrt{6}\approx 1.5505 \\
& x=4+ \sqrt{6} \approx 6.4495
\end{aligned}
\end{equation} Horizontal intercepts: $(1.5505 ,0),(6.4495,0)$.
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: $[-18,\infty)$
