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{{ printedBook.courseTrack.name }} {{ printedBook.name }} To determine the domain and range, we will draw the graph of the given quadratic function. Note that it is written in standard form. Let's identify the values of $a,$ $b,$ and $c.$ $\begin{gathered} y=\dfrac{2}{3}x^2 +12 \quad\Leftrightarrow\quad y=\textcolor{#ff8c00}{\dfrac{2}{3}}x^2+\textcolor{#9400D3}{0}x+\textcolor{#ff00ff}{12} \end{gathered}$ We can see that $\textcolor{#ff8c00}{a}=\textcolor{#ff8c00}{\frac{2}{3}},$ $\textcolor{#9400D3}{b}=\textcolor{#9400D3}{0},$ and $\textcolor{#ff00ff}{c}=\textcolor{#ff00ff}{12}.$ Since $\textcolor{#9400D3}{b}=\textcolor{#9400D3}{0},$ we can say that the given quadratic function is in the form $y=ax^2 +c.$ Now, we will follow four steps to graph the function.

- Find the axis of symmetry.
- Determine the vertex.
- Make a table of values and identify their reflections across the axis of symmetry.
- Connect the points with a parabola.

Recall that the given function is in the form $y=ax^2 +c.$ Therefore, the axis of symmetry of the parabola is the $y\text{-}$axis.

The value of $\textcolor{#ff00ff}{c}$ tells us that the parabola has the same shape as $y=\textcolor{#ff8c00}{\frac{2}{3}}x^2$ but it is translated $\textcolor{#ff00ff}{12}$ units up. Since the vertex of $y=\textcolor{#ff8c00}{\frac{2}{3}}x^2$ is at the origin, the vertex of $y=\textcolor{#ff8c00}{\frac{2}{3}}x^2+\textcolor{#ff00ff}{12}$ is at $(0,\textcolor{#ff00ff}{12}).$

To finish the graph we need at least one point on each side of the axis of symmetry. Let's find the values of $y$ when $x=1$ and $x=3.$

$x$ | $\dfrac{2}{3}x^2+12$ | $y=\dfrac{2}{3}x^2+12$ |
---|---|---|

${\color{#0000FF}{1}}$ | $\dfrac{2}{3}({\color{#0000FF}{1}})^2+12$ | $12\frac{2}{3}$ |

${\color{#0000FF}{3}}$ | $\dfrac{2}{3}({\color{#0000FF}{3}})^2+12$ | $18$ |

Let's plot these points, $\left(1,12\frac{2}{3}\right)$ and $(3,18),$ and their reflections across the axis of symmetry.

We can now draw the graph of the function. Since $\textcolor{#ff8c00}{a}=\textcolor{#ff8c00}{\frac{2}{3}},$ which is positive, the parabola will open upwards. Let's connect the points with a smooth curve.

There are no restrictions for the input, so the domain is all real numbers. We know that the maximum or minimum point on a parabola is the vertex. Since the parabola opens upwards the vertex is the minimum point. Therefore, all possible outputs must be *greater than or equal to* $\textcolor{#ff00ff}{12}.$
$\begin{aligned}
\textbf{Domain:} \ & \text{all real numbers}\\
\textbf{Range:} \ & y\geq 12
\end{aligned}$