Answer
A) vertex :$ (-1,2)$
B) Vertical intercept: $(0,3)$,
Horizontal intercept: None
C) See the graph
D) Domain: All real numbers.
Range: Range: $[2,\infty )$
Work Step by Step
Given \begin{equation}\begin{aligned}
g(x)& =x^2+2x+3\\
a& = 1,\quad b= 2, \quad c= 3
\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)+3\\
& =2.
\end{aligned}
\end{equation} The vertex of the function is $(-1,2)$.
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& = 3.
\end{aligned}
\end{equation} Vertical intercept: $(0,3)$.
2) Find the horizontal intercept by setting the function to zero and solving for $x$.\begin{equation}
\begin{aligned}
&x^2+2 x+3=0 \\
a& = 1,\quad b= 2, \quad c= 3\\
x&= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
x& =\frac{-(2) \pm \sqrt{(2)^2-4(1)(3)}}{2(1)} \\
& =\frac{-2 \pm \sqrt{-8}}{2}.
\end{aligned}
\end{equation} The equation has no solution, therefore there is no horizontal intercept.
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: $[2,\infty )$
