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Study Guides > MATH 0123

The Graph of a Quadratic Function

Learning Outcomes

  • Graph quadratic functions using tables and transformations
  • Identify important features of the graph of a quadratic function of the form f(x)=ax2+bx+cf(x)=ax^2+bx+c
Quadratic functions form parabolas when graphed in the coordinate plane, so you can be sure fairly quickly when graphing a quadratic if you have the correct shape. As with lines in the plane, creating a table of input and output values then plotting points will reveal the shape. But unlike straight lines between points, the parabola is a smooth curve.  Later in this module, you'll learn some other good methods for sketching a quick, accurate graph of a quadratic equation. It is helpful to have an idea about what the shape of the graph of a quadratic function should be so you can be sure that you have chosen enough points to plot as a guide when sketching the graph. Let's start with the most basic quadratic function, f(x)=x2f(x)=x^{2}. Graph f(x)=x2f(x)=x^{2}. Start with a table of values. Then think of each row of the table as an ordered pair.
x f(x)
2−2 44
1−1 11
00 00
11 11
22 44
Plot the points (2,4),(1,1),(0,0),(1,1),(2,4)(-2,4), (-1,1), (0,0), (1,1), (2,4) Graph with the point negative 2, 4; the point negative 1, 1; the point 0, 0; the point 1,1; the point 2,4. Since the points are not on a line, you cannot use a straight edge. Connect the points as best as you can using a smooth curve (not a series of straight lines). You may want to find and plot additional points (such as the ones in blue here). Placing arrows on the tips of the lines implies that they continue in that direction forever. A curved U-shaped line through the points from the previous graph. Notice that the shape is similar to the letter U. This is called a parabola. One-half of the parabola is a mirror image of the other half. The lowest point on this graph is called the vertex. The vertical line that goes through the vertex is called the line of reflection. In this case, that line is the y-axis. The equations for quadratic functions have the form f(x)=ax2+bx+cf(x)=ax^{2}+bx+c where a0 a\ne 0. In the basic graph above, a=1a=1, b=0b=0, and c=0c=0. In the following video, we show an example of plotting a quadratic function using a table of values. https://youtu.be/wYfEzOJugS8 Changing a changes the width of the parabola and whether it opens up (a>0a>0) or down (a<0a<0). If a is positive, the vertex is the lowest point, if a is negative, the vertex is the highest point. In the following example, we show how changing the value of a will affect the graph of the function.

Example

Match each function with its graph. a) f(x)=3x2 \displaystyle f(x)=3{{x}^{2}} b) f(x)=3x2 \displaystyle f(x)=-3{{x}^{2}} c) f(x)=12x2 \displaystyle f(x)=\frac{1}{2}{{x}^{2}} 1) compared to g(x)=x squared 2) compared to g(x)=x squared 3) compared to g(x)=x squared

Answer: Function a) matches graph 22 Function b) matches graph 11 Function c) matches graph 33   Function a) f(x)=3x2 \displaystyle f(x)=3{{x}^{2}} means that inputs are squared and then multiplied by three, so the outputs will be greater than they would have been for f(x)=x2f(x)=x^2.  This results in a parabola that has been squeezed, so graph 22 is the best match for this function. compared to g(x)=x squared Function b) f(x)=3x2 \displaystyle f(x)=-3{{x}^{2}} means that inputs are squared and then multiplied by negative three, so the outputs will be farther away from the xx-axis than they would have been for f(x)=x2f(x)=x^2, but negative in value, so graph 11 is the best match for this function. compared to g(x)=x squared Function c) f(x)=12x2 \displaystyle f(x)=\frac{1}{2}{{x}^{2}} means that inputs are squared then multiplied by 12\dfrac{1}{2}, so the outputs are less than they would be for f(x)=x2f(x)=x^2.  This results in a parabola that has been opened wider thanf(x)=x2f(x)=x^2. Graph 33 is the best match for this function. compared to g(x)=x squared

If there is no b term, changing c moves the parabola up or down so that the y intercept is (0,c0, c). In the next example, we show how changes to affect the graph of the function.

Example

Match each of the following functions with its graph. a) f(x)=x2+3 \displaystyle f(x)={{x}^{2}}+3 b) f(x)=x23 \displaystyle f(x)={{x}^{2}}-3   1) compared to g(x)=x squared 2) compared to g(x)=x squared

Answer: Function a) f(x)=x2+3 \displaystyle f(x)={{x}^{2}}+3 means square the inputs then add three, so every output will be moved up 33 units. The graph that matches this function best is 22. compared to g(x)=x squared Function b) f(x)=x23 \displaystyle f(x)={{x}^{2}}-3  means square the inputs then subtract three, so every output will be moved down 33 units. The graph that matches this function best is 11. compared to g(x)=x squared

Changing bb moves the line of reflection, which is the vertical line that passes through the vertex ( the high or low point) of the parabola. It may help to know how to calculate the vertex of a parabola to understand how changing the value of bb in a function will change its graph. To find the vertex of the parabola, use the formula (b2a,f(b2a)) \displaystyle \left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right). For example, if the function being considered is f(x)=2x23x+4f(x)=2x^2-3x+4, to find the vertex, first calculate b2a\Large\frac{-b}{2a} a=2a = 2, and b=3b = -3, therefore b2a=(3)2(2)=34\dfrac{-b}{2a}=\dfrac{-(-3)}{2(2)}=\dfrac{3}{4}. This is the xx value of the vertex. Now evaluate the function at x=34x =\Large\frac{3}{4} to get the corresponding y-value for the vertex. f(b2a)=2(34)23(34)+4=2(916)94+4=181694+4=9894+4=98188+328=238f\left( \dfrac{-b}{2a} \right)=2\left(\dfrac{3}{4}\right)^2-3\left(\dfrac{3}{4}\right)+4=2\left(\dfrac{9}{16}\right)-\dfrac{9}{4}+4=\dfrac{18}{16}-\dfrac{9}{4}+4=\dfrac{9}{8}-\dfrac{9}{4}+4=\dfrac{9}{8}-\dfrac{18}{8}+\dfrac{32}{8}=\dfrac{23}{8}. The vertex is at the point (34,238)\left(\dfrac{3}{4},\dfrac{23}{8}\right).  This means that the vertical line of reflection passes through this point as well.  It is not easy to tell how changing the values for bb will change the graph of a quadratic function, but if you find the vertex, you can tell how the graph will change. In the next example, we show how changing can change the graph of the quadratic function.

Example

Match each of the following functions with its graph. a) f(x)=x2+2x \displaystyle f(x)={{x}^{2}}+2x b) f(x)=x22x \displaystyle f(x)={{x}^{2}}-2x a) compared to g(x)=x squared b) compared to g(x)=x squared

Answer: Find the vertex of function a) f(x)=x2+2x \displaystyle f(x)={{x}^{2}}+2x. a=1,b=2a = 1, b = 2 x-value: b2a=22(1)=1\dfrac{-b}{2a}=\dfrac{-2}{2(1)}=-1 y-value: f(b2a)=(1)2+2(1)=12=1f(\dfrac{-b}{2a})=(-1)^2+2(-1)=1-2=-1. Vertex = (1,1)(-1,-1), which means the graph that best fits this function is a) compared to g(x)=x squared   Find the vertex of function b) f(x)=x22xf(x)={{x}^{2}}-2x. a=1,b=2a = 1, b = -2 x-value: b2a=22(1)=1\dfrac{-b}{2a}=\dfrac{2}{2(1)}=1 y-value: f(b2a)=(1)22(1)=12=1f(\dfrac{-b}{2a})=(1)^2-2(1)=1-2=-1. Vertex = (1,1)(1,-1), which means the graph that best fits this function is b) compared to g(x)=x squared

Note that the vertex can change if the value for c changes because the y-value of the vertex is calculated by substituting the x-value into the function. Here is a summary of how the changes to the values for a, b, and, c of a quadratic function can change it is graph.

Properties of a Parabola

For f(x)=ax2+bx+c \displaystyle f(x)=a{{x}^{2}}+bx+c, where a, b, and c are real numbers,
  • The parabola opens upward if a>0a > 0 and downward if a<0a < 0.
  • a changes the width of the parabola. The parabola gets narrower if a>1|a|> 1 and wider if a<1|a|<1.
  • The vertex depends on the values of a, b, and c. The vertex is (b2a,f(b2a))\left(\dfrac{-b}{2a},f\left( \dfrac{-b}{2a}\right)\right).
In the last example, we showed how you can use the properties of a parabola to help you make a graph without having to calculate an exhaustive table of values.

Example

Graph f(x)=2x2+3x3f(x)=−2x^{2}+3x–3.

Answer: Before making a table of values, look at the values of a and c to get a general idea of what the graph should look like. a=2a=−2, so the graph will open down and be thinner than f(x)=x2f(x)=x^{2}. c=3c=−3, so it will move to intersect the y-axis at (0,3)(0,−3). To find the vertex of the parabola, use the formula (b2a,f(b2a)) \displaystyle \left( \dfrac{-b}{2a},f\left( \dfrac{-b}{2a} \right) \right). Finding the vertex may make graphing the parabola easier.

Vertex formula=(b2a,f(b2a))\text{Vertex }\text{formula}=\left( \dfrac{-b}{2a},f\left( \dfrac{-b}{2a} \right) \right)

x-coordinate of vertex:

b2a=(3)2(2)=34=34 \displaystyle \dfrac{-b}{2a}=\dfrac{-(3)}{2(-2)}=\dfrac{-3}{-4}=\dfrac{3}{4}

y-coordinate of vertex:

f(b2a)=f(34)   f(34)=2(34)2+3(34)3                  =2(916)+943                  =1816+943                  =98+188248                  =158 \displaystyle \begin{array}{l}f\left( \dfrac{-b}{2a} \right)=f\left( \dfrac{3}{4} \right)\\\,\,\,f\left( \dfrac{3}{4} \right)=-2{{\left( \dfrac{3}{4} \right)}^{2}}+3\left( \dfrac{3}{4} \right)-3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-2\left( \dfrac{9}{16} \right)+\dfrac{9}{4}-3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{-18}{16}+\dfrac{9}{4}-3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{-9}{8}+\dfrac{18}{8}-\dfrac{24}{8}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\dfrac{15}{8}\end{array}

Vertex: (34,158) \displaystyle \left( \dfrac{3}{4},-\dfrac{15}{8} \right) Use the vertex, (34,158) \displaystyle \left( \dfrac{3}{4},-\dfrac{15}{8} \right), and the properties you described to get a general idea of the shape of the graph. You can create a table of values to verify your graph. Notice that in this table, the x values increase. The y values increase and then start to decrease again. This indicates a parabola.
x f(x)
2−2 17−17
1−1 8−8
00 3−3
11 2−2
22 5−5
Vertex at negative three-fourths, negative 15-eighths. Other points are plotted: the point negative 2, negative 17; the point negative 1, negative 8; the point 0, negative 3; the point 1, negative 2; and the point 2, negative 5. Connect the points as best you can using a smooth curve. Remember that the parabola is two mirror images, so if your points do not have pairs with the same value, you may want to include additional points (such as the ones in blue shown below). Plot points on either side of the vertex. x=12x=\Large\frac{1}{2} and x=32x=\Large\frac{3}{2} are good values to include. A parabola drawn through the points in the previous graph

The following video shows another example of plotting a quadratic function using the vertex. https://youtu.be/leYhH_-3rVo  

Summary

Creating a graph of a function is one way to understand the relationship between the inputs and outputs of that function. Creating a graph can be done by choosing values for x, finding the corresponding y values, and plotting them. However, it helps to understand the basic shape of the function. Knowing how changes to the basic function equation affect the graph is also helpful. The shape of a quadratic function is a parabola. Parabolas have the equation f(x)=ax2+bx+cf(x)=ax^{2}+bx+c, where a, b, and c are real numbers and a0a\ne0. The value of a determines the width and the direction of the parabola, while the vertex depends on the values of a, b, and c. The vertex is (b2a,f(b2a)) \displaystyle \left( \dfrac{-b}{2a},f\left( \dfrac{-b}{2a} \right) \right).

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