Answer
A) Vertex : $(-7 ,-15.25)$
B) Vertical intercept: $(0,-27.5)$
Horizontal intercept: None
C) See the graph
D) Domain: All real numbers.
Range:$(-\infty, -15.25] $
Work Step by Step
Given \begin{equation}
\begin{aligned}
M(x) &=-0.25 x^2-3.5 x-27.5\\
a & =-0.25, \quad b=-3.5, \quad c=-27.5
\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 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}{2 a}=-\frac{-3.5}{2\cdot(-0.25)}=-7 \\
& \begin{aligned}
M(-7) & =-0.5(-7)^2-3.5(-7)-27.5 \\
& =-15.25
\end{aligned}
\end{aligned}
\end{equation} The vertex of the function is $(-7 ,-15.25)$.
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& =-27.5
\end{aligned}
\end{equation} Vertical intercept: $(0,-27.5)$.
2) Find the horizontal intercept by setting the function to zero and solve. \begin{equation}
\begin{aligned}
\frac{-0.25 x^2-3.5 x-27.5}{-0.25}& =0 \\
x^2+14x+110&=0
\end{aligned}
\end{equation} Determine $x$: \begin{equation}
\begin{aligned}
x & =\frac{-14 \pm \sqrt{(14)^2-4(1)(110)}}{2 \cdot(1)} \\
& =-\frac{-14 \pm \sqrt{-224}}{2} \\
\end{aligned}
\end{equation} No solution.
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:$(-\infty, -15.25] $
