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Identifying Characteristics of Quadratic Functions

Identifying Characteristics of Quadratic Functions 1.11 - Solution

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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,a, b,b, and c.c. y=23x2+12y=23x2+0x+12\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 a=23,\textcolor{#ff8c00}{a}=\textcolor{#ff8c00}{\frac{2}{3}}, b=0,\textcolor{#9400D3}{b}=\textcolor{#9400D3}{0}, and c=12.\textcolor{#ff00ff}{c}=\textcolor{#ff00ff}{12}. Since b=0,\textcolor{#9400D3}{b}=\textcolor{#9400D3}{0}, we can say that the given quadratic function is in the form y=ax2+c.y=ax^2 +c. Now, we will follow four steps to graph the function.

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

Finding the Axis of Symmetry

Recall that the given function is in the form y=ax2+c.y=ax^2 +c. Therefore, the axis of symmetry of the parabola is the y-y\text{-}axis.

Determining the Vertex

The value of c\textcolor{#ff00ff}{c} tells us that the parabola has the same shape as y=23x2y=\textcolor{#ff8c00}{\frac{2}{3}}x^2 but it is translated 12\textcolor{#ff00ff}{12} units up. Since the vertex of y=23x2y=\textcolor{#ff8c00}{\frac{2}{3}}x^2 is at the origin, the vertex of y=23x2+12y=\textcolor{#ff8c00}{\frac{2}{3}}x^2+\textcolor{#ff00ff}{12} is at (0,12).(0,\textcolor{#ff00ff}{12}).

Making the Table of Values

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

xx 23x2+12\dfrac{2}{3}x^2+12 y=23x2+12y=\dfrac{2}{3}x^2+12
1{\color{#0000FF}{1}} 23(1)2+12\dfrac{2}{3}({\color{#0000FF}{1}})^2+12 122312\frac{2}{3}
3{\color{#0000FF}{3}} 23(3)2+12\dfrac{2}{3}({\color{#0000FF}{3}})^2+12 1818

Let's plot these points, (1,1223)\left(1,12\frac{2}{3}\right) and (3,18),(3,18), and their reflections across the axis of symmetry.

Connecting the Points

We can now draw the graph of the function. Since a=23,\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.

Domain and Range

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 12.\textcolor{#ff00ff}{12}. Domain: all real numbersRange: y12\begin{aligned} \textbf{Domain:} \ & \text{all real numbers}\\ \textbf{Range:} \ & y\geq 12 \end{aligned}