Answer
A) Vertex :$ (5.5,6.25)$
B) Vertical intercept: $(0,-24)$,
Horizontal intercepts: $(3,0),(8,0)$
C) See the graph
D) Domain: All real numbers.
Range: $(-\infty, 6.25] $
Work Step by Step
Given\begin{equation}
\begin{aligned}
m(b)&=-b^2+11 b-24\\
a& = -1,\quad d= 11, \quad c= -24.
\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{-d}{2a} \\
& =\frac{-(11)}{2(-1)} \\
& = 5.5 \\
m(5.5)&=-1 \cdot (5.5)^2+11\cdot (5.5)-24\\
& =6.25
\end{aligned}
\end{equation} The vertex of the function is $(5.5,6.25)$.
Part B)
Step 3: Find the vertical and horizontal intercepts.
1) Find the vertical intercept which equal to the constant $c$ if the parabola is in its standard form.
\begin{equation}
\begin{aligned}
y& = -24
\end{aligned}
\end{equation} Vertical intercept: $(0,-24)$.
2) Find the horizontal intercept by setting the function to zero and solving for $x$. \begin{equation}
\begin{aligned}
&-b^2+11 b-24=0 \\
a& = -1,\quad d= 11, \quad c= -24\\
x&=\frac{-b\pm\sqrt{d^2-4ac}}{2a}\\
b&=\frac{-(11) \pm \sqrt{(11)^2-4(-1)(-24)}}{2(-1)} \\
& =\frac{-11 \pm \sqrt{25}}{-2}\\
& = -\frac{-11\pm 5}{2}\\
&= -(-5.5\pm 2.5)\\
&
x&=-(-5.5- 2.5)= 8 \\
x&=-(-5.5+ 2.5)= 3.
\end{aligned}
\end{equation}
Horizontal intercepts: $(3,0) , (8,0)$.
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, 6.25] $
