Answer
1) Vertex: $(-5,7)$
2) Vertical intercept: $(0,14.5)$
Horizontal intercept: None
3) Domain: All real numbers.
Range: $[7, \infty) $
Work Step by Step
Given \begin{equation}
\begin{aligned}
h(n)&=0.3 n^2+3 n+14.5.
\end{aligned}
\end{equation} We determine the constants $a$, $b$, $c$: $$\begin{align*}
h(n)&=an^2+bn+c\\
a& = 0.3, b= 3, c= 14.5.
\end{align*}$$ 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.
Step 2: Find the vertex. This function is in standard form for a quadratic, so we can use the formula: \begin{equation}
\begin{aligned}
n & =\frac{-b}{2 a} \\
& =\frac{-( 3)}{2(0.3)} \\
& =-5\\
h(-5) & =0.3\cdot (-5)^2+3\cdot (-5)+14.5 \\
& =7.
\end{aligned}
\end{equation} The vertex of the function is $(-5,7)$.
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}
y& = 14.5.
\end{aligned}
\end{equation} Vertical intercept: $(0,14.5)$.
2) Find the horizontal intercept by setting the function to zero and solve.
\begin{equation}
\begin{aligned}
\left(0.3 n^2+3 n+14.5 \right)\cdot (10)& =0\cdot (10) \\
3 x^2+30 x+145 & =0 \\
x & =\frac{-(30)\pm \sqrt{(30)^2-4\cdot(3)\cdot(145)}}{2\cdot(3)} \\
& = \frac{-30\pm \sqrt{-840}}{6}.
\end{aligned}
\end{equation} No solution. There is no horizontal intercept.
The graph of the function is shown in the figure with the required points shown.
