Answer
A) Vertex: $(-8 ,-9.5)$
B) Vertical intercept: $(0,6.5)$
Horizontal intercepts: $(-14.164,0),(-1.806,0)$
C) See the graph
D) Domain: All real numbers.
Range: $[-9.5,\infty )$
Work Step by Step
Given \begin{equation}
\begin{aligned}
f(x)&=0.25 x^2+4 x+6.5\\
a& = 0.25 ,\quad b= 4, \quad c= 6.5.
\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 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}
& x=\frac{-b}{2 a}=\frac{-4}{2(0.25)}=-8 \\
& \begin{aligned}
f(4.5) & =0.25(-8)^2+4(-8)+6.5 \\
& =-9.5.
\end{aligned}
\end{aligned}
\end{equation} The vertex of the function is $(-8 ,-9.5)$.
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& = 6.5
\end{aligned}
\end{equation}
Vertical intercept: $(0,6.5)$.
2) Find the horizontal intercept by setting the function to zero and solve. \begin{equation}
\begin{aligned}
\frac{\left(0.25 x^2+4 x+6.5\right) \cdot 10}{0.25} & =\frac{0.10}{0.25} \\
x^2+16 x+26 & =0
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
x & =\frac{-16 \pm \sqrt{16^2-4(1)(26)}}{2 \cdot(1)} \\
& =\frac{-16 \pm \sqrt{152}}{2} \\
& =\frac{-16 \pm 12.38828}{2}
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
& x=\frac{-16-12.38828}{2}=-14.164 \\
& x=\frac{-16+12.38828}{2}=-1.806
\end{aligned}
\end{equation}
Horizontal intercepts: $(-14.164,0),(-1.806,0)$.
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: $[-9.5,\infty )$
