Answer
A) Vertex :$(-6,79)$
B) Vertical intercept: $(0,43)$,
Horizontal intercepts: $(-14.89,0) , (2.89,0)$
C) See the graph
D) Domain: All real numbers.
Range: Range: $(-\infty, 79]$
Work Step by Step
Given\begin{equation}
\begin{aligned}
D(z)&=-z^2-12 z+43\\
a& = -1,\quad b= -12, \quad c= 43\\
\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{-(-12)}{2(-1)} \\
& =-6\\
D(-6) & =-(-6)^2-12(-6)+43 \\
& =79.
\end{aligned}
\end{equation} The vertex of the function is $(-6,79)$.
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}
z& = 43.
\end{aligned}
\end{equation} Vertical intercept: $(0,43)$.
2) Find the horizontal intercept by setting the function to zero and solving for $x$.
\begin{equation}
\begin{aligned}
& -z^2-12 z+43&==0 \\
a& = -1,\quad b= -12 \quad c= 43\\
z&=\frac{-(-12) \pm \sqrt{(-12)^2-4(-1)(43)}}{2(-1)} \\
z&=-\frac{12 \pm \sqrt{316}}{2} \\
& z\approx-(6 \pm 8.888)\\
& z=-(6-8.888)=2.888 \\
& z=-(6+8888)=-14.888
\end{aligned}
\end{equation} Horizontal intercepts: $(-14.89,0), (2.89,0)$
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: $(-\infty, 79]$
