Answer
A) Vertex: $(-3.75,-73.25)$
B) Vertical intercept: $(0,-17)$
Horizontal intercepts: $(-8.03,0), (0.53,0)$
C) See the graph
D) Domain: All real numbers.
Range: $[-73.25.5,\infty )$
Work Step by Step
Given
\begin{equation}
\begin{aligned}
h(t)&=4 t^2+30 t-17\\
a& = 4,\quad b= 30, \quad c= -17.
\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 up since $a$ is positive.
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}
t & =\frac{-b}{2 a} \\
& =\frac{-(30)}{2(4)} \\
& =-3.75\\
h(-3.75) & =4(-3.75)^2+30(-3.75)-17 \\
& =-73.25
\end{aligned}
\end{equation} The vertex of the function is $(-3.75,-73.25)$.
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}
h& = -17.
\end{aligned}
\end{equation} Vertical intercept: $(0,-17)$.
2) Find the horizontal intercept by setting the function to zero and solve.
\begin{equation}
\begin{aligned}
&4 t^2+30 t-17= 0 \\
t & =\frac{-30 \pm \sqrt{30^2-4 \cdot 4 \cdot(-17)}}{2 \cdot 4} \\
& =\frac{-30 \pm \sqrt{1172}}{8} \\
& \approx -3.75 \pm 4.28\\
& t=-3.75-4.28=-8.03 \\
& t=-3.75+4.28=0.53.
\end{aligned}
\end{equation} Horizontal intercepts: $(-8.03,0) , (0.53,0)$
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:$[-73.25,\infty )$
