Answer
A) Vertex: $(-1.75,2.8125)$
B) Vertical intercept: $(0,-12.5)$
Horizontal intercept: $(-2.5, 0),(-1, 0)$
C) See the graph
D) Domain: All real numbers.
Range: $(-\infty, 2.8125]$
Work Step by Step
\begin{equation}
\begin{aligned}
P(k&)=-5 k^2-17.5 k-12.5\\
a& = -5 ,\quad b= -17.5, \quad , c= -12.5\\
\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}
k&=\frac{b}{2a}=-\frac{(-17.5)}{2\cdot (-5)}=-1.75 \\
P(-1.75) & =-5(-1.75)^2-17.5(-1.75)-12.5 \\
& =2.8125.
\end{aligned}
\end{equation} The vertex of the function is $(-1.75,2.8125)$.
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& = -12.5
\end{aligned}
\end{equation} Vertical intercept: $(0,-12.5)$.
2) Find the horizontal intercept by setting the function to zero and solve. \begin{equation}
\begin{aligned}
-5 k^2-17.5 k-12.5 & =0 \\
5 k^2+17.5 k+12.5 & =0
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
k&=\frac{-(17.5) \pm \sqrt{(17.5)^2-4(5)(12.5)}}{2 \cdot 5}\\
k&=\frac{-17.5 \pm 7.5}{10}\\
k&=-1.75 \pm 0.75
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
& k=-1.75-0.75=-2.5 \\
& k=-1.75+0.75=-1.
\end{aligned}
\end{equation}
Horizontal intercept $(-2.5, 0),(-1, 0)$.
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: $(-\infty, 2.8125]$
