Answer
A) Vertex: $(-2.4333,6.5633)$
B) Vertical intercept: $(0,-11.2)$
Horizontal intercepts: $(-3.912,0),(-0.954,0)$
C) See the graph
D) Domain: All real numbers.
Range: $(-\infty, 6.5633]$
Work Step by Step
Given \begin{equation}
\begin{aligned}
f(x)&=-3 x^2-14.6 x-11.2\\
a &=-3 ,\quad b= -14.6 , \quad c= -11.2.
\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 down since $a$ is negative.
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}{2a}=-\frac{(-14.6)}{2\cdot (-3)}=-2.4333 \\
P(-1.75) & =-3(-2.4333 )^2-14.6(-2.4333 )-11.2 \\
& =6.5633.
\end{aligned}
\end{equation}
The vertex of the function is $(-2.4333,6.5633)$.
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}
y& = -11.2.
\end{aligned}
\end{equation} Vertical intercept: $(0,-11.2)$.
2) Find the horizontal intercept by setting the function to zero and solve. \begin{equation}
\begin{aligned}
-3 x^2-14.6 x-11.2 & =0 \\
3 x^2+14.6 x+11.2 & =0
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
& x=\frac{-14.6 \pm \sqrt{14.6^2-4(3)(11.2)}}{2(3)} \\
& =\frac{-141.6 \pm \sqrt{78.76}}{6} \\
& =-2.4633 \pm 1.479
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
& x=-2.4633-1.479=-3.912 \\
& x=-2.4633+1.479=-0.954.
\end{aligned}
\end{equation} Horizontal intercepts: $(-3.912,0),(-0.954,0)$
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: $(-\infty, 6.5633]$
