Answer
A) Vertex: $(3 ,1)$
B) Vertical intercept: $(0,4)$
Horizontal intercept: None
C) See the graph
D) Domain: All real numbers.
Range: $[1,\infty )$
Work Step by Step
\begin{equation}
\begin{aligned}
h(t)&=\frac{1}{3} t^2-2 t+4\\
a & =\frac{1}{3}, \quad b=-2, \quad c=4.
\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}
& t=\frac{-b}{2 a}=-\frac{-2}{2\cdot(\frac{1}{3})}=3 \\
& \begin{aligned}
h(3) & =\frac{1}{3}(3)^2-2(3)+4 \\
& =1.
\end{aligned}
\end{aligned}
\end{equation} The vertex of the function is $(3 ,1)$.
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}
y& = 4.
\end{aligned}
\end{equation} Vertical intercept: $(0,4)$.
2) Find the horizontal intercept by setting the function to zero and solve. \begin{equation}
\begin{aligned}
\frac{1}{3} t^2-2 t+4 & =0 \\
t^2-6 t+12 & =0
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
t & =\frac{-(-6) \pm \sqrt{(-6)^2-4(1)(12)}}{2 \cdot(1)} \\
& =\frac{-16 \pm \sqrt{-12}}{2}.
\end{aligned}
\end{equation} The equation has no solution.
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: $[1,\infty )$
