Lesson Explainer: Graphing Quadratic Functions | Nagwa Lesson Explainer: Graphing Quadratic Functions | Nagwa

Lesson Explainer: Graphing Quadratic Functions Mathematics • Third Year of Preparatory School

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In this explainer, we will learn how to graph any quadratic function using a table of values and a given interval and identify the features of the graph.

Quadratic equations are used in everyday life, in science, business, and engineering. They can determine the paths of moving objects, from bouncing balls to the flight path of a projectile. Businesses can use them to forecast revenues and design packaging to minimize waste. We can use quadratic equations to identify the minimum and maximum values of many different variables such as speed, cost, and area. This means we can use quadratic equations to analyze these scenarios. In particular, we can use the graphs of quadratic functions to easily determine information about the maximum and minimum values and when the outputs will be equal to zero.

To do this, let’s start by recalling what we mean by a quadratic function: it is a single-variable degree-two polynomial function. This means that quadratic functions have the form 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, for some constants π‘Ž,𝑏, and 𝑐, where π‘Ž is nonzero. A quadratic equation is an equation in the form 𝑦=𝑓(π‘₯), where 𝑓(π‘₯) is a quadratic function.

We can better understand quadratics by considering their graphs. For example, let’s consider the graphs of the two quadratic equations 𝑦=π‘₯+4 and 𝑦=βˆ’π‘₯+9.

We note that these graphs have a very similar shape, known as a parabola. In fact, all quadratics have this shape, and the sign of the constant π‘Ž tells us if the parabola opens upward or downward. If π‘Ž>0, then the parabola opens upward, and if π‘Ž<0, then the parabola opens downward. We can see this in the two given graphs; the first graph has a positive leading coefficient and its graph opens upward, while the second has a negative leading coefficient and its graph opens downward.

There are also many useful points we can label to help us determine and convey information about the graph.

First, we recall that the 𝑦-intercept of the graph will be the point on the graph where π‘₯=0, which is where the curve crosses the 𝑦-axis. In the first graph, this is (0,4), and in the second, this is (0,9). These tell us the outputs of the quadratic when we input π‘₯=0. It is worth noting that we can find the 𝑦-intercept of the graph of the function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨ by substituting in π‘₯=0. We have 𝑓(0)=π‘ŽΓ—0+𝑏×0+𝑐=π‘οŠ¨. So, it has 𝑦-intercept (0,𝑐).

Second, the π‘₯-intercepts of the graph will be the points on the graph where 𝑦=0, which is where the curve crosses the π‘₯-axis. We can see that the first curve has no π‘₯-intercepts and the second has two at the points (βˆ’3,0) and (3,0). The π‘₯-coordinates of these points tell us the input values at which the function outputs 0.

It is worth noting that since the 𝑦-coordinates of the π‘₯-intercepts are 0, we often only talk about the π‘₯-coordinates of these points. For example, we could say that the π‘₯-intercepts of the second graph are at βˆ’3 and 3. In a similar way, we could say that the 𝑦-intercept of the second graph is at 9.

In our first example, we will determine the π‘₯-intercepts of a quadratic curve from its graph.

Example 1: Identifying the π‘₯-Intercepts of a Quadratic from Its Graph

Identify the π‘₯-intercepts of the quadratic function 𝑓(π‘₯)=π‘₯βˆ’1.

Answer

We recall that the intercepts of a graph are the π‘₯-coordinates of the points where the graph intercepts the π‘₯-axis. In other words, they are the points on the curve with the 𝑦-coordinate equal to 0. We can see that there are two such points and we can mark them on our diagram.

Hence, the π‘₯-intercepts are π‘₯=βˆ’1 and π‘₯=1.

We can see that points such as the π‘₯- and 𝑦-intercepts can help us to determine features of the graph of a quadratic function. One way of sketching a quadratic graph is to use a table of values. For example, let’s say we want to sketch 𝑦=βˆ’π‘₯+9, where βˆ’3≀π‘₯≀3. We construct a table of input values of π‘₯ and we calculate the corresponding outputs of 𝑓(π‘₯)=βˆ’π‘₯+9.

We can calculate 𝑓(βˆ’3) by substituting π‘₯=βˆ’3 into the function as follows (and remember to be careful when squaring negative numbers): 𝑓(βˆ’3)=βˆ’(βˆ’3)+9=βˆ’9+9=0.

We follow the same process for all of the integers in the given interval to get the following table.

π‘₯βˆ’3βˆ’2βˆ’10123
𝑓(π‘₯)0589850

Each column of the table gives us the coordinates of a point on the graph of 𝑦=βˆ’π‘₯+9. We can plot these points onto a coordinate grid and then sketch a parabola through the points to sketch the curve 𝑦=βˆ’π‘₯+9.

An interesting property we can notice about this graph is that the curve is symmetric in the 𝑦-axis. We can also see this fact by noting the outputs in the function table are symmetric about π‘₯=0. Every parabola has a vertical line of symmetry (called its axis of symmetry); however, it is not always the line π‘₯=0.

To see an example of this, consider the following graph of 𝑦=βˆ’π‘₯+4π‘₯+12. We can label its π‘₯-intercepts, 𝑦-intercept, and line of symmetry as shown.

In this case, the parabola is symmetric in the line π‘₯=2. We can see that this line of symmetry passes through the point where the parabola turns. This point is called the vertex, and it will tell us the maximum or minimum output of the function depending on the sign of π‘Ž. We can note that since parabolas are symmetric in the vertical line through their axis of symmetry, this line must lie halfway between the π‘₯-intercepts (if the parabola has π‘₯-intercepts). In fact, it will lie halfway between any π‘₯-values with the same output.

We can summarize all of these properties of quadratic functions and their graphs as follows.

Properties: Quadratic Functions and Their Graphs

  • Quadratic functions have the form 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, for some constants π‘Ž,𝑏, and 𝑐, where π‘Ž is nonzero.
  • All quadratic curves have a parabolic shape. If π‘Ž>0, then the graph of the quadratic will open upward; if π‘Ž<0, then the graph of the quadratic will open downward.
  • All graphs of quadratics have a 𝑦-intercept at the point (0,𝑐); this is the point where the graph crosses the 𝑦-axis.
  • Graphs of quadratics can have 0, 1, or 2 π‘₯-intercepts. These are the points the graph crosses the π‘₯-axis at. The π‘₯-coordinates of these points tell us the input values for which the function outputs 0.
  • All graphs of quadratics have a single turning point called the vertex. The sign of π‘Ž tells us if the turning point will be a maximum or a minimum.
  • All parabolas are symmetric in the vertical line through their vertex.
    The axis of symmetry lies halfway between the π‘₯-intercepts (if the parabola has π‘₯-intercepts).

In our next example, we will identify the axis of symmetry for a given quadratic curve.

Example 2: Identifying the Axis of Symmetry for a Given Quadratic Function

Identify the axis of symmetry of the quadratic function 𝑓(π‘₯)=π‘₯βˆ’1.

Answer

We begin by recalling that the axis of symmetry for all quadratic curves is the vertical line passing through its vertex. We also recall that the vertex of a quadratic is often referred to as its turning point.

We can find the coordinates of this point from the given graph or by recalling that the axis of symmetry will lie halfway between the π‘₯-intercepts.

We see in the diagram that the π‘₯-intercepts are at βˆ’1 and 1. The halfway point of these values is βˆ’1+12=0.

Hence, the axis of symmetry is the line π‘₯=0. We can also note that the vertex is the point of intersection between the curve and axis of symmetry, which has coordinates (0,βˆ’1).

Earlier, we demonstrated how we can use a table of values (function table) to help us sketch the graph of a quadratic that was symmetric about the 𝑦-axis. We can, in fact, use a similar method for any quadratic function. In particular, we want to find the coordinates of its vertex and any intercepts (if possible).

For example, if we wanted to sketch the graph of 𝑓(π‘₯)=π‘₯βˆ’π‘₯βˆ’6, we could evaluate the function at a set of values of π‘₯ to determine the coordinates of points on the curve. Let’s try the integers from βˆ’2 to 2.

We see that 𝑓(βˆ’2)=(βˆ’2)βˆ’(βˆ’2)βˆ’6=4+2βˆ’6=0.

We can evaluate the rest of the integers in a similar manner to get the following table.

π‘₯βˆ’2βˆ’1012
𝑓(π‘₯)0βˆ’4βˆ’6βˆ’6βˆ’4

We want to sketch the graph using this table, and to do this, we want to find the coordinates of the vertex of the graph. By looking at the table, we can see that there is a repeated output of βˆ’6:

π‘₯βˆ’2βˆ’1012
𝑓(π‘₯)0βˆ’4βˆ’6βˆ’6βˆ’4

This means that the vertex lies midway between these points on the line π‘₯=12.

We can find the coordinates of the vertex by evaluating 𝑓(π‘₯) at π‘₯=12. We have 𝑓12=ο€Ό12οˆβˆ’12βˆ’6=βˆ’254.

Hence, the coordinates of the vertex are (0.5,βˆ’6.25).

We can then plot these five points on the coordinate plane and connect the points with a parabola. We get the following sketch.

Let’s now see some examples of sketching quadratics by using a table of values.

Example 3: Sketching a Quadratic and Finding Its Intercepts and Vertex

  1. By completing the table of values for 𝑓(π‘₯)=(π‘₯+2)βˆ’4, identify the correct graph of the quadratic function on the domain [βˆ’4,0].
    π‘₯βˆ’4βˆ’3βˆ’2βˆ’10
    𝑓(π‘₯)
  2. Identify the π‘₯-intercepts of the given quadratic function.
  3. Identify the 𝑦-intercept of the given quadratic function.
  4. Identify the axis of symmetry of the given quadratic function.
  5. Identify the vertex of the given quadratic function.

Answer

Part 1

We begin by filling in the table of values. To do this, we need to evaluate the function at the given values of π‘₯; we have 𝑓(βˆ’4)=(βˆ’4+2)βˆ’4=0,𝑓(βˆ’3)=(βˆ’3+2)βˆ’4=βˆ’3,𝑓(βˆ’2)=(βˆ’2+2)βˆ’4=βˆ’4,𝑓(βˆ’1)=(βˆ’1+2)βˆ’4=βˆ’3,𝑓(0)=(0+2)βˆ’4=0.

We can then fill in the table of values.

π‘₯βˆ’4βˆ’3βˆ’2βˆ’10
𝑓(π‘₯)0βˆ’3βˆ’4βˆ’30

Each column then gives us the π‘₯- and 𝑦-coordinates of a point that lies on the graph. We can see that 𝑓(βˆ’3)=𝑓(βˆ’1) and that 𝑓(βˆ’4)=𝑓(0), so the axis of symmetry will be between these π‘₯-values.

We can sketch these points on the coordinate plane and connect them with a parabolic shape to sketch the graph, where we note that the domain of the function is [βˆ’4,0] so we do not sketch the graph of the function for π‘₯-values outside of this range.

This is the graph given in option E.

Part 2

There are two methods we can use to determine the π‘₯-intercepts. First, we can see the π‘₯-intercepts in the graph are (βˆ’4,0) and (0,0). Second, we can use the table to see that 𝑓(βˆ’4)=0 and 𝑓(0)=0.

Using either method, we can see the π‘₯-intercepts are π‘₯=βˆ’4 and π‘₯=0.

Part 3

We can see that the 𝑦-intercept is also (0,0) from the graph. Alternatively, we can recall that the 𝑦-intercept is 𝑓(0)=0.

In either case, the 𝑦-intercept is 𝑦=0.

Part 4

We recall the axis of symmetry is a vertical line of symmetry for the parabola. We can identify the axis of symmetry from the graph.

The parabola is symmetric in the line π‘₯=βˆ’2. It is also worth noting we can see this in the table.

We see that 𝑓(βˆ’3)=βˆ’3 and 𝑓(βˆ’1)=βˆ’3; these have the same output value, so the line of symmetry must be in the middle of the inputs. Since (βˆ’3)+(βˆ’1)2=βˆ’2, the axis of symmetry is π‘₯=βˆ’2.

Alternatively, we could have used the fact that the line of symmetry will be halfway between the π‘₯-intercepts: βˆ’4+02=βˆ’2.

Thus, the axis of symmetry is the line π‘₯=βˆ’2.

Part 5

The vertex of the parabola is the turning point. We can see this in the graph or, alternatively, we can use the fact that the vertex lies on the axis of symmetry.

We see that the vertex has coordinates (βˆ’2,βˆ’4).

In our final two examples, we will identify the correct sketch of a quadratic curve by constructing a table of values.

Example 4: Identifying the Graph of a Quadratic Curve

Which of the following graphs represents the equation 𝑦=π‘₯βˆ’5π‘₯+8 plotted for 0≀π‘₯≀5?

Answer

We can sketch the graph of a quadratic by recalling that the coordinates of any point on the graph of a function 𝑓(π‘₯) have the form (π‘₯,𝑓(π‘₯)). This means we can sketch the graph of a function by evaluating the function at various input values. We do this by filling in a table.

We say that 𝑓(π‘₯)=π‘₯βˆ’5π‘₯+8 and we have 𝑓(0)=(0)βˆ’5(0)+8=8,𝑓(1)=(1)βˆ’5(1)+8=4,𝑓(2)=(2)βˆ’5(2)+8=2,𝑓(3)=(3)βˆ’5(3)+8=2,𝑓(4)=(4)βˆ’5(4)+8=4,𝑓(5)=(5)βˆ’5(5)+8=8.

This gives the following table.

π‘₯012345
𝑓(π‘₯)842248

We can note that 𝑓(2)=𝑓(3). So, the line of symmetry of the curve will be halfway between the π‘₯-values at 2.5.

Hence, the line of symmetry of the parabola is π‘₯=2.5. We can determine the coordinates of the vertex by substituting π‘₯=52 into the function to get 𝑓(2.5)=(2.5)βˆ’5(2.5)+8=1.75.

Thus, the vertex of the parabola is (2.5,1.75).

We can now sketch a parabola connecting these points, where it is important to note that we are sketching our function to have π‘₯-values between 0 and 5. This gives us the following.

This is the graph given in option C.

It is worth noting that we can check our answer (or eliminate the other options) by using the properties of the graphs of quadratics. For example, we see that the leading coefficient of the given quadratic is 1. Since this is positive, we know that the parabola must open upward.

Let’s see another example of identifying the correct graph of a quadratic.

Example 5: Identifying the Graph of a Quadratic Curve

Which of the following graphs represents the equation 𝑦=βˆ’2π‘₯+8π‘₯βˆ’6 plotted for 0≀π‘₯≀4?

Answer

There are several different ways we could answer this question. For example, we could eliminate options by using the properties of quadratic curves and the given equation. However, we will sketch a graph of the quadratic and then determine which option matches our sketch.

To do this, we recall that the coordinates of any point on the graph of a function 𝑓(π‘₯) have the form (π‘₯,𝑓(π‘₯)). This means we can sketch the graph of a function by evaluating the function at various input values. We do this by filling in a table. We say that 𝑓(π‘₯)=βˆ’2π‘₯+8π‘₯βˆ’6 and we have 𝑓(0)=βˆ’2(0)+8(0)βˆ’6=βˆ’6,𝑓(1)=βˆ’2(1)+8(1)βˆ’6=0,𝑓(2)=βˆ’2(2)+8(2)βˆ’6=2,𝑓(3)=βˆ’2(3)+8(3)βˆ’6=0,𝑓(4)=βˆ’2(4)+8(4)βˆ’6=βˆ’6.

This gives the following table.

π‘₯01234
𝑓(π‘₯)βˆ’6020βˆ’6

We can note that the outputs of the function are symmetric about π‘₯=2, so the vertex of the parabola is π‘₯=2.

We can then sketch a parabola connecting these five points, where it is important to note that we are sketching our function to have π‘₯-values between 0 and 4. This gives us the following.

This is the graph given in option A.

It is worth noting that we can check our answer (or eliminate the other options) by using the properties of the graphs of quadratics. For example, we see that the leading coefficient of the given quadratic is βˆ’2. Since this is negative, we know that the parabola must open downward.

Let’s finish by recapping some of the important points of this explainer.

Key Points

  • A quadratic equation is an equation of the form 𝑦=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, where π‘Ž,𝑏, and 𝑐 are constants and π‘Žβ‰ 0. This is called the standard form.
  • All quadratics have the shape of a parabola that opens upward when π‘Ž is positive and opens downward when π‘Ž is negative.
  • All graphs of quadratics have a 𝑦-intercept at the point (0,𝑐); this is the point where the graph crosses the 𝑦-axis.
  • Graphs of quadratics can have 0, 1, or 2 π‘₯-intercepts. These are the points the graph crosses the π‘₯-axis at. The π‘₯-coordinates of these points tell us the input values for which the function outputs 0.
  • All graphs of quadratics have a single turning point called the vertex. The sign of π‘Ž tells us whether the turning points of these graphs will be maxima or minima.
  • All parabolas are symmetric in the vertical line through their vertex.
    The axis of symmetry lies halfway between the π‘₯-intercepts (if the parabola has π‘₯-intercepts).

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