Lesson Explainer: Graphing Simple Quadratic Functions Mathematics

In this explainer, we will learn how to graph and interpret quadratic functions of the form 𝑦=π‘˜π‘₯+π‘οŠ¨.

The word quadratus is Latin and means β€œto square.” In mathematics, the term quadratic is derived from this and describes something that relates to squares, squaring, or equations that involve terms where the variable is raised to the power of 2.

In particular, a quadratic function is a function of the form 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, where π‘Ž, 𝑏, and 𝑐 are real numbers and π‘Žβ‰ 0. In this explainer, we will primarily be focusing on functions in which 𝑏=0, meaning our equations will be of the form 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨.

With this in mind, let’s recap how to work with functions.

Definition: Functions and Completing Function Tables

When a relationship assigns exactly one output for a given input, it is called a function. Since π‘₯ is the input to the function, the value of the function for a certain number can be found by substituting that number for the variable π‘₯ in the function equation. This process can be repeated any number of times and organized in a function table.

π‘₯π‘₯π‘₯π‘₯οŠ©β‹―π‘₯
𝑓(π‘₯)𝑓(π‘₯)οŠ§π‘“(π‘₯)οŠ¨π‘“(π‘₯)οŠ©β‹―π‘“(π‘₯)

To create a table of values for a function of the form 𝑦=π‘˜π‘₯+π‘οŠ¨, we substitute values of π‘₯ into the function equation and fully simplify the result. We can then plot the resulting ordered pairs on a coordinate plane.

In our first example, we will demonstrate this process in more detail.

Example 1: Completing a Table of Values for a Simple Quadratic Function

This is a table for 𝑓(π‘₯)=π‘₯+2. Complete it by finding the values of π‘Ž, 𝑏, 𝑐, and 𝑑.

π‘₯βˆ’2βˆ’1012
𝑓(π‘₯)π‘Žπ‘2𝑐𝑑

Answer

Remember, to complete a table of values for a function of the form 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨, we substitute each value of π‘₯ into the function.

So, to find the value of π‘Ž, we substitute π‘₯=βˆ’2 into the function 𝑓(π‘₯)=π‘₯+2: π‘Ž=𝑓(βˆ’2)=(βˆ’2)+2=4+2=6.

Next, to find the value of 𝑏, we let π‘₯=βˆ’1: 𝑏=𝑓(βˆ’1)=(βˆ’1)+2=1+2=3.

To find the value of 𝑐, we let π‘₯=1: 𝑐=𝑓(1)=1+2=1+2=3.

Finally, we substitute π‘₯=2 to find the value of 𝑑: 𝑑=𝑓(2)=2+2=4+2=6.

Let’s check our method by calculating 𝑓(0) and checking that our method gives the correct output of 2: 𝑓(0)=0+2=0+2=2.

Since the value of 𝑓(0) corresponds to the value given in the table, we can be fairly confident about our method. The table of values can be completed as follows.

π‘₯βˆ’2βˆ’1012
𝑓(π‘₯)63236

So, the correct values are π‘Ž=6, 𝑏=3, 𝑐=3, and 𝑑=6.

In the previous example, we could have calculated the value of 𝑓(βˆ’1) by substituting π‘₯=βˆ’1 into the function: 𝑓(βˆ’1)=(βˆ’1)+2=3.

Adding this value to the table, we now have a completed set of ordered pairs that can be used to draw the graph of the function.

π‘₯βˆ’2βˆ’1012
𝑓(π‘₯)63236

Let’s begin by plotting the pairs of coordinates, as demonstrated on the diagram below.

Since the values of the function do not change linearly, we do not join these points with a straight line. In fact, for polynomial functions of order of two and above the points should be connected with a smooth curve.

The shape of this completed graph is a symmetrical curve called a parabola. In the special case of quadratic functions of the form 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨, the line of symmetry is, in fact, the 𝑦-axis, or the line π‘₯=0. The same cannot necessarily be said for more complicated quadratic functions of the form 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, although the line of symmetry will always pass through the vertex (turning point) of the graph of a quadratic function. Note that, for simple quadratic functions, the symmetry of the curve can be observed in the tables of values, which gives us a helpful way of checking our results.

In our next example, we will demonstrate how to complete a table of values for a quadratic function and then sketch its graph.

Example 2: Completing a Table of Values for a Simple Quadratic Function and Using this to Find the Graph of the Function

  1. Complete the following table for the graph of 𝑓(π‘₯)=2βˆ’2π‘₯ by finding the values of π‘Ž, 𝑏, 𝑐, and 𝑑.
    π‘₯βˆ’2βˆ’1012
    𝑓(π‘₯)π‘Žπ‘π‘π‘‘βˆ’6
  2. Which figure represents the graph of the function 𝑓(π‘₯)?

Answer

Part 1

Remember, to complete a table of values for a function of the form 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨, we substitute each value of π‘₯ into the function. The equation 𝑓(π‘₯)=2βˆ’2π‘₯ might look different from this but is in fact of the same form with the terms in a different order.

First, let π‘₯=βˆ’2 to find the value of π‘Ž, remembering to apply the order of operations: π‘Ž=𝑓(βˆ’2)=2βˆ’2(βˆ’2)=2βˆ’2Γ—4=2βˆ’8=βˆ’6.

Next, letting π‘₯=βˆ’1, 𝑏=𝑓(βˆ’1)=2βˆ’2(βˆ’1)=2βˆ’2Γ—1=2βˆ’2=0.

The value of 𝑐 is found by substituting π‘₯=0: 𝑐=𝑓(0)=2βˆ’2(0)=2βˆ’2Γ—0=2βˆ’0=2.

Finally, we find 𝑑 by calculating 𝑓(1): 𝑑=𝑓(1)=2βˆ’2(1)=2βˆ’2Γ—1=2βˆ’2=0.

The table of values for 𝑓(π‘₯)=2βˆ’2π‘₯ is as follows.

π‘₯βˆ’2βˆ’1012
𝑓(π‘₯)βˆ’6020βˆ’6

Hence, π‘Ž=βˆ’6, 𝑏=0, 𝑐=2, and 𝑑=0.

Part 2

To find the graph of 𝑓(π‘₯), we can write the values from the table as a list of ordered pairs of the form (π‘₯,𝑓(π‘₯)).

These are (βˆ’2,βˆ’6), (βˆ’1,0), (0,2), (1,0), and (2,βˆ’6). The graph of 𝑓(π‘₯) will be a smooth curve passing through all of these points.

This is option A.

In our previous example, the function 𝑓(π‘₯)=2βˆ’2π‘₯ generated an β€œinverted,” or upside down, parabola. Let’s compare this with the graph of the function 𝑔(π‘₯)=2+2π‘₯.

While the graphs of these functions share the same 𝑦-intercept, the parabola is n shaped when the coefficient of π‘₯ is negative and u shaped when the coefficient of π‘₯ is positive. This means that the vertex of the function 𝑓(π‘₯) corresponds to the maximum value of the function, while the vertex of 𝑔(π‘₯) is the minimum.

Let’s demonstrate this in another example.

Example 3: Identifying the Graph of a Simple Quadratic Function

Which of the following graphs represents the quadratic function 𝑓(π‘₯)=π‘₯+0.5 on the interval [βˆ’2,2]?

Answer

To draw the graph of a function 𝑓(π‘₯), we can begin by constructing a table listing the values of π‘₯ and 𝑓(π‘₯). Since each graph is given over the interval βˆ’2≀π‘₯≀2, we will calculate the values of 𝑓(π‘₯) over this interval.

π‘₯βˆ’2βˆ’1012
𝑓(π‘₯)

To find the first entry in the second row of this table, we calculate 𝑓(βˆ’2) by substituting π‘₯=βˆ’2 into the expression π‘₯+0.5: 𝑓(βˆ’2)=(βˆ’2)+0.5=4+0.5=4.5.

Next, we calculate 𝑓(βˆ’1) by substituting π‘₯=βˆ’1: 𝑓(βˆ’1)=(βˆ’1)+0.5=1+0.5=1.5.

Continuing in this manner, 𝑓(0)=0+0.5=0.5,οŠ¨π‘“(1)=1+0.5=1.5,οŠ¨π‘“(2)=2+0.5=4.5.

So, the completed table of values for 𝑓(π‘₯)=π‘₯+0.5 is as follows.

π‘₯βˆ’2βˆ’1012
𝑓(π‘₯)4.51.50.51.54.5

The ordered pairs that we will plot on a pair of axes are (βˆ’2,4.5), (βˆ’1,1.5), (0,0.5), (1,1.5), and (2,4.5). Since these satisfy a quadratic function, we will join them with a smooth curve as shown in the diagram.

This is graph B.

So far, we have sketched the graph of a quadratic function of the form 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨ by using tables to create ordered pairs (π‘₯,𝑓(π‘₯)). We have seen that graphs of this form are symmetric parabolas and that, in the special case of quadratic functions of the form 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨, the line of symmetry is the line π‘₯=0.

We can deduce one further property of these functions. Recall that the 𝑦-intercept of a function 𝑦=𝑓(π‘₯) is found by substituting π‘₯=0 and solving the resulting equation: 𝑓(0)=π‘˜Γ—0+𝑐=𝑐.

Hence, the 𝑦-intercept for the graph of 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨ is 𝑐; it has coordinates (0,𝑐). For equations of this form, the 𝑦-intercept is also the location of the vertex of the parabola. This is not generally true, however, for more complicated quadratic functions.

In our next example, we will demonstrate how to use these properties to identify the correct graph for a quadratic function.

Example 4: Identifying Graphs of Simple Quadratic Functions and Commenting on Their Differences

  1. Which graph represents the quadratic function 𝑓(π‘₯)=π‘₯+3?
  2. Which graph represents the quadratic function 𝑓(π‘₯)=π‘₯+4?
  3. Which of the following is true about the two graphs?
    1. The two curves have the same shape but the second is a vertical shift of the first.
    2. The two curves have the same shape but the second is a horizontal shift of the first.
    3. The first curve is just a stretched form of the second curve.
    4. The two curves are identical.
    5. One curve is obtained by rotating the other by 90∘ about the origin.

Answer

Part 1

Remember, graphs of the form 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨ are symmetric parabolas with a line of symmetry π‘₯=0. The 𝑦-intercept for the graph of 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨ is 𝑐. Hence, the graph of 𝑓(π‘₯)=π‘₯+3 is a parabola with a line of symmetry given by π‘₯=0 and a 𝑦-intercept of 3. The only graph with a 𝑦-intercept of 3 is graph D.

We can check this answer by calculating some coordinates of points on the curve. For instance, let’s evaluate 𝑓(βˆ’1) and 𝑓(2): 𝑓(βˆ’1)=(βˆ’1)+3=4,οŠ¨π‘“(2)=2+3=7.

So, the points (βˆ’1,4) and (2,7) must lie on the curve of 𝑓(π‘₯)=π‘₯+3.

The graph that satisfies these criteria is D.

Part 2

The graph of 𝑓(π‘₯)=π‘₯+4 is a parabola with a line of symmetry π‘₯=0 and a 𝑦-intercept of 4. The only graph that satisfies these criteria is graph C. Let’s check by calculating coordinates that lie on the curve; we can evaluate 𝑓(βˆ’2) and 𝑓(1): 𝑓(βˆ’2)=(βˆ’2)+4=8,οŠ¨π‘“(1)=1+4=5.

The points (βˆ’2,8) and (1,5) lie on the curve 𝑓(π‘₯)=π‘₯+4.

This is option C.

Part 3

We can answer this question by considering the properties of graphs of the form 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨ alongside the graphs of our functions. Let’s redefine those as 𝑓(π‘₯)=π‘₯+3 and 𝑓(π‘₯)=π‘₯+4. These graphs are symmetrical parabolas with a line of symmetry about π‘₯=0 and 𝑦-intercepts of 3 and 4 respectively. We observe that the graphs of 𝑓(π‘₯)=π‘₯+3 and 𝑓(π‘₯)=π‘₯+4 have the same general shape but different 𝑦-intercepts.

Hence, the answer is option A. The two graphs have the same shape, but the second is a vertical shift of the first.

In our previous example, we saw the effect of varying the constant 𝑐 in the equation 𝑦=π‘˜π‘₯+π‘οŠ¨. Varying 𝑐 results in a vertical shift of the function either up or down. For example, increasing 𝑐 by one will result in a shift of the whole graph vertically upward by a factor of one, whereas decreasing 𝑐 by one would result in a shift of the whole graph vertically downward.

What happens then if we vary π‘˜? The first thing to note, as we recalled in our commentary following example 2, is that the sign of π‘˜ affects the shape of the parabola: a positive π‘˜ results in a quadratic that resembles a u shape and a negative π‘˜ results in a quadratic that resembles an n shape. If we consider the value of π‘˜ between the two, that is, when π‘˜=0, we have the constant function 𝑦=𝑐. For reference, let’s include the graphs of 𝑦=π‘˜π‘₯+1 for values of π‘˜ equal to βˆ’2, 0, and 2.

If we think about other values of π‘˜, starting with those between βˆ’2 and 0, then we will have a more complete idea of how varying π‘˜ affects the graph of the quadratic.

Let’s plot the graph of 𝑦=βˆ’π‘₯+1.

Here, we can see that the parabola still resembles an n shape, whose vertex has the coordinates (0,1), but the slopes of the graph on either side of the vertex change more slowly when compared with the function 𝑦=βˆ’2π‘₯+1. Visually, we can see that the width of the β€œn shape” increased when we increased the value of π‘˜ from βˆ’2 to βˆ’1. If we further increased the value of π‘˜ from βˆ’1 to βˆ’0.5, we would see that the width of the graph would further increase.

This behavior continues in a similar way for positive values of π‘˜, but here, an increase in the value of π‘˜ would result in a parabola whose slopes on either side of the vertex change more quickly, that is, a β€œu shape” that becomes increasingly thin.

In summary, we can conclude that the larger the absolute value of π‘˜ is, the faster the slope of the parabola increases or decreases.

In our final example, we will demonstrate how to identify the equation of a quadratic function 𝑦=π‘˜π‘₯+π‘οŠ¨ given its graph.

Example 5: Determining a Quadratic Equation from its Graph

Which of the following is the equation of the function drawn on the graph?

  1. 𝑓(π‘₯)=βˆ’π‘₯+8
  2. 𝑓(π‘₯)=π‘₯βˆ’8
  3. 𝑓(π‘₯)=βˆ’π‘₯βˆ’8
  4. 𝑓(π‘₯)=π‘₯+8
  5. 𝑓(π‘₯)=π‘₯βˆ’8

Answer

We begin by observing the general shape of the given graph. It is a parabola, meaning that it is the graph of a quadratic function, and it has a line of symmetry given by the 𝑦-axis. This means its equation will be of the form 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨, where π‘˜β‰ 0. Our job now will be to identify the values of π‘˜ and 𝑐.

Since the value of the 𝑦-intercept is found by calculating 𝑓(0), and the 𝑦-intercept of our graph is βˆ’8, 𝑓(0)=π‘˜Γ—0+𝑐=𝑐=βˆ’8.

Hence, 𝑐=βˆ’8, meaning that the equation can be written as 𝑓(π‘₯)=π‘˜π‘₯βˆ’8. Next, we can calculate the value of π‘˜ by choosing a point that lies on the given curve. Let’s choose (3,1). This tells us that 𝑓(3)=1. Substituting this equation in place of 𝑓(3), we can form and solve an equation for π‘˜: 1=π‘˜Γ—3βˆ’81=9π‘˜βˆ’89π‘˜=9π‘˜=1.

In fact, we would expect π‘˜>0 since the parabola is u shaped.

The equation of the curve is 𝑓(π‘₯)=π‘₯βˆ’8, which is option B.

We will now finish by recapping the key concepts from this explainer.

Key Points

  • To create a table of values for a function, we substitute values of π‘₯ into the function. We can then plot the resulting ordered pairs on a coordinate plane.
  • The graphs of quadratic functions are symmetrical about a vertical line that passes through their vertex.
  • The shape of a quadratic curve is called a parabola. When the coefficient of π‘₯ is positive, the graph is u shaped, and when it is negative, the graph is n shaped. The magnitude, or absolute value of the coefficient of π‘₯, affects how quickly the slope of the parabola increases or decreases.
  • Graphs of the form 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨ are symmetric parabolas with a line of symmetry π‘₯=0.
  • The 𝑦-intercept for the graph of 𝑓(π‘₯)=π‘˜π‘₯+π‘οŠ¨ is located at its vertex and has coordinates (0,𝑐). Therefore, increasing or decreasing the value of 𝑐 results in a vertical shift of the parabola either upward or downward.

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