Answer
1) Vertex :$(2,28)$.
2) Vertical intercept: $(0,16)$
Horizontal intercepts: $(-1.0551,0), (5.0551,0)$
3) Domain: All real numbers
Range: $(-\infty, 28] $
Work Step by Step
Given \begin{equation}
\begin{aligned}
f(x)&=-3 x^2+12 x+16.
\end{aligned}
\end{equation} We determine the constants $a$, $b$, $c$: $$\begin{align*}
f(x)&=ax^2+bx+c\\
a& = -3, b= 12, c= 16.
\end{align*}$$ 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.
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{-( 12)}{2(-3)} \\
& =2\\
f(2) & =-3\cdot (2)^2+12\cdot (2)+16 \\
& =28.
\end{aligned}
\end{equation} The vertex of the function is $(2,28)$.
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}
y& = 16
\end{aligned}
\end{equation} Vertical intercept: $(0,16)$.
2) Find the horizontal intercept by setting the function to zero and solve.
\begin{equation}
\begin{aligned}
\left(-3 x^2+12 x+16 \right)\cdot (-1)& =0\cdot (-1) \\
3 x^2-12 x-16 & =0 \\
x & =\frac{-(-12)\pm \sqrt{(-12)^2-4\cdot(3)\cdot(-16)}}{2\cdot(3)} \\
& = \frac{12\pm \sqrt{336}}{6} \\
x & =2 \pm 3.0551\\
x_1& = 2-3.0551= -1.0551\\
x_2& = 2+3.0551= 5.0551.
\end{aligned}
\end{equation} The graph of the function is shown in the figure with the required points shown.
