Answer
A) Vertex :$ (-1,-16)$
B) Vertical intercept: $(0,-15)$,
Horizontal intercept $(-5,0)\quad (3,0)$
C) See graph
D) Domain: All real numbers.
Range: Range: $[-16,\infty )$
Work Step by Step
Given\begin{equation}
\begin{aligned}
f(x) & =x^2+2 x-15\\
a& = 1,\quad b= 2, \quad c= -15\\
\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}{2a} \\
& =\frac{-(2)}{2(1)} \\
& = -1 \\
f(-1)&=1 \cdot (-1)^2+2\cdot (-1)-15\\
& =-16.
\end{aligned}
\end{equation} The vertex of the function is $(-1,-16)$.
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& = -15.
\end{aligned}
\end{equation}Vertical intercept: $(0,-15)$.
2) Find the horizontal intercept by setting the function to zero and solving for $x$.
\begin{equation}
\begin{aligned}
x^2+2 x-15& =0 \\
a& = 1,\quad b= 2, \quad c= -15\\
x&= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
x& =\frac{-(2) \pm \sqrt{(2)^2-4(1)(-15)}}{2(1)} \\
& =\frac{-2 \pm \sqrt{64}}{2}\\
&= \frac{-2\pm 8}{2}\\
&= -1\pm 4\\
& x=-1- 4= -5
& x=-1+ 4= 3
\end{aligned}
\end{equation} 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: $[-16,\infty )$
