Answer
A) Vertex: $(0,-5)$
B) Vertical intercept: $(0,-5)$
Horizontal intercept: None
C) See the graph
D) Domain: All real numbers.
Range: $(-\infty, -5] $
Work Step by Step
Given \begin{equation}
\begin{aligned}
h(x)&=-\frac{1}{4} x^2-5\\
a& = -\frac{1}{4} ,\quad b= 0, \quad c= -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}
x & =\frac{-b}{2 a} \\
& =\frac{-(0)}{2\cdot (-1/4)} \\
& =0\\
h(0) & =-5 \\
& =-5.
\end{aligned}
\end{equation} The vertex of the function is $(0,-5)$.
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}
h& = -5
\end{aligned}
\end{equation} Vertical intercept: $(0,-5)$.
2) Find the horizontal intercept by setting the function to zero and solve.
\begin{equation}
\begin{aligned}
-\frac{1}{4} x^2-5 & =0 \\
-\frac{1}{4} x^2 & =5 \\
x^2 & =-20.
\end{aligned}
\end{equation} This equation has no solution. Hence, there is no horizontal intercept.
Horizontal intercept: None
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: $(-\infty ,-5]$
