Lesson Explainer: Solving Quadratic Equations Using Graphs | Nagwa Lesson Explainer: Solving Quadratic Equations Using Graphs | Nagwa

Lesson Explainer: Solving Quadratic Equations Using Graphs Mathematics

In this explainer, we will learn how to graph a quadratic function of the form 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, in order to solve the equation 𝑓(π‘₯)=0.

When working with any particular quadratic function, it can be helpful to remind ourselves of some basic properties that these functions hold, as well as the key features that can be used to classify or describe them. Consider the general quadratic function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+𝑐, where π‘Ž, 𝑏, and 𝑐 are real numbers and π‘Žβ‰ 0. When plotting this function, there will be two basic shapes that we will see, depending on the sign of the parameter π‘Ž. If π‘Ž is positive, then the plot of the function will have a β€œu” shape, and if π‘Ž is negative, then the plot of the function will have an β€œn” shape. We demonstrate this on the graphs below, where the left-hand graph is the plot of a quadratic function where π‘Ž is positive, and the right-hand graph is the plot of a quadratic function where π‘Ž is negative. We have deliberately omitted the π‘₯𝑦-axis, as well as the gridlines.

The values of the other two parameters, 𝑏 and 𝑐, will of course affect the graph when plotting a quadratic function, and one way that this is understood is by classifying how all parameters influence the roots of the function, which we will focus on later. However, the main focus of this explainer will be to understand how roots of a quadratic function can be found by plotting the graph. This actually splits the problem into three distinct categories, each of which we will give a short example of. Before doing so, we will give a definition of the so-called solution set of a quadratic, as follows.

Definition: The Solution Set of a Quadratic Function

Consider the quadratic function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+𝑐, where π‘Ž, 𝑏, and 𝑐 are real numbers and π‘Žβ‰ 0. Then the solution set 𝑆 of the function 𝑓(π‘₯) consists of all possible roots of 𝑓(π‘₯). In other words, 𝑆 is the set of all values of π‘₯ such that 𝑓(π‘₯)=0, which is where the graph intersects the π‘₯-axis. There are three distinct possibilities for the solution set:

  • There are two real roots of 𝑓(π‘₯), when π‘₯=π‘ οŠ§ and π‘₯=π‘ οŠ¨, where π‘ β‰ π‘ οŠ§οŠ¨. In this case, the solution set 𝑆 has two elements and is written 𝑆={𝑠,𝑠}.
  • There is one real root of 𝑓(π‘₯), when π‘₯=π‘ οŠ§. In this case, the solution set has one element and is written 𝑆={𝑠}.
  • There are no real roots of 𝑓(π‘₯), meaning that the solution set is empty. In this case, we say that 𝑆 is equal to the β€œempty set” which is denoted 𝑆=βˆ….

Consider the quadratic function 𝑓(π‘₯)=π‘₯βˆ’π‘₯βˆ’2. One excellent way of understanding the behavior of this function (or any function) is to plot a graph or a high-quality sketch. We can begin doing this by plotting a table of values and then converting this information to coordinates in the π‘₯𝑦-plane. For the quadratic function 𝑓(π‘₯)=π‘₯βˆ’π‘₯βˆ’2, we currently have no knowledge of how this behaves and therefore we do not have any information as to where interesting behavior might occur (such as the location of the roots). With no better guess at hand, we will create a table of values for the few points near to the π‘₯-axis. In particular, we will create a table of values for π‘₯=βˆ’3,βˆ’2,βˆ’1,0,1,2,3, which in this case gives the following.

π‘₯βˆ’3βˆ’2βˆ’10123
𝑓(π‘₯)1040βˆ’2βˆ’204

Now that we have this information about the function, we can plot seven coordinates to better gain an understanding of its behavior. We would begin with the coordinate (βˆ’3,10), then with (βˆ’2,4) and so on until we had obtained the plot below.

This information alone offers a much better understanding of how the function behaves. Either from the table of values above, or from the plotted coordinates, we can see that there are two points where the function meets with the π‘₯-axis. It is perhaps clearer to first plot the full quadratic function in this region, which we have done below.

In the graph above, we have marked the points where the function crosses the π‘₯-axis in red crosses. This appears to occur in two locations: when π‘₯=βˆ’1 and π‘₯=2. Equally, we could say that the coordinates of the roots are (βˆ’1,0) and (2,0). Finally, we could use the language of the solution set as described in the definition above, and we would say the solution set is 𝑆={βˆ’1,2}. If we wished to, we could check that these were the correct roots by factoring 𝑓(π‘₯), which in this case would return 𝑓(π‘₯)=(π‘₯+1)(π‘₯βˆ’2). The roots of the function occur when 𝑓(π‘₯)=0, which is precisely when π‘₯+1=0 or when π‘₯βˆ’2=0; that is, when π‘₯=βˆ’1 or π‘₯=2.

Now, we will give an example of a quadratic function where there is only one root, and we will see how this invokes a curious property that will appear several times again in this explainer. Suppose that we now had the quadratic function 𝑓(π‘₯)=4π‘₯βˆ’32π‘₯+64. A superb first step toward understanding this function is with the following table of values.

π‘₯2345678
𝑓(π‘₯)16404163664

Unlike the previous example, every value of 𝑓(π‘₯) is zero or positive, which might hint that there is a fundamentally different behavior occurring. In the graph below, we have plotted the table of values as well as the quadratic function within this region. We have had to scale the axis in order to properly demonstrate the behavior around the π‘₯-axis, which is where any roots will occur.

There is clearly one root when π‘₯=4, which is marked with a red cross. Alternatively, we could say that the single root has the coordinate (4,0), or that the solution set is 𝑆={4}. Notice how the single root is also the minimum point of the quadratic function. This is an instance of a more general result: when the solution set has one element, the vertex maximum or minimum is also the only root of the function. Had we chosen to factor the quadratic then, we would have found that 𝑓(π‘₯)=(2π‘₯βˆ’8) and then 𝑓(π‘₯)=0 occurs only when π‘₯=4.

For our final example, we will consider the quadratic function 𝑓(π‘₯)=βˆ’π‘₯βˆ’3π‘₯βˆ’5, which produces the table of values below. We should note that all given values of 𝑓(π‘₯) are negative and, crucially, there are none that are zero, which is a clue that the behavior is different from the previous example.

π‘₯βˆ’3βˆ’2βˆ’10123
𝑓(π‘₯)βˆ’5βˆ’3βˆ’3βˆ’5βˆ’9βˆ’15βˆ’23

We have plotted this function below and can see immediately that there are no points where the function intersects the π‘₯-axis, meaning that there are no roots of the function. Given this, we would write the solution set as 𝑆=βˆ…. Had we attempted to factor this quadratic, either by completing the square or through other means, then we would have found that this was not possible as, at some point, we would have been attempting to take the square root of a negative number. We will comment more on this later in the explainer.

Now we have seen an example of solution set with two elements, a solution set with one element, and a solution set with zero elements. There are various relationships between parameters that can be established which determine critical behaviors of quadratic functions. We will allude to several such examples throughout the rest of this explainer, but briefly we will reconsider the function plotted immediately above. Notice that the vertex (the maximum point) is below the π‘₯-axis. Given that the plot of this quadratic is an β€œn” shape, this means that there cannot be any points where the function intersects the π‘₯-axis because, by definition, a function cannot ever exceed its maximum value. It is possible to argue such results with total algebraic rigor, which we will leave as an exercise for the interest of the reader.

We could also be asked a question where we are required to plot the graph of a quadratic and then use this to estimate its roots (and therefore the solution set). As ever, it is generally good practice to plot or sketch the graph of any function, no matter what the intended purpose of working with this function actually is. Often, this will give a strong indication as to the behavior of the function and the nature of any specific features (such as roots and turning points).

Let us consider the function 𝑓(π‘₯)=3π‘₯βˆ’3π‘₯βˆ’18 in the interval π‘₯=βˆ’4 to π‘₯=4 for integer values of π‘₯. By plotting the graph of the function, we can then use this to estimate the solutions to equation 𝑓(π‘₯)=0. We create the following table of values for the function.

π‘₯βˆ’4βˆ’3βˆ’2βˆ’101234
𝑓(π‘₯)42180βˆ’12βˆ’18βˆ’18βˆ’12018

We can use this to plot the graph as follows.

In order to best demonstrate the features around the line 𝑓(π‘₯)=0, we can scale the axes accordingly. As we can see, this graph seems to strongly indicate that the the roots of 𝑓(π‘₯) are found when π‘₯=βˆ’2 and π‘₯=3, which would imply that the solution set can be written as 𝑆={βˆ’2,3}. In fact, in this case, it is obvious from the table of values, since we clearly see that 𝑓(βˆ’2)=𝑓(3)=0.

In our first formal example, we will enjoy the luxury of having the graph preplotted for our convenience. This will not always be the case, and in real life we would often have to plot such a curve manually or, even better, use an online graph platform to gain a more precise (and speedier) understanding.

Example 1: Finding the Solution Set of a Quadratic Equation Graphically

The diagram shows the graph of 𝑦=𝑓(π‘₯). What is the solution set of the equation 𝑓(π‘₯)=0?

Answer

Recall that the solutions to the equation 𝑓(π‘₯)=0 correspond to the π‘₯-values of the points on the graph of 𝑦=𝑓(π‘₯) where the curve intersects the π‘₯-axis. This is because 𝑦=0 at these points.

In the graph, we can see that this happens at exactly one point, at (βˆ’2,0), which is also the location of the minimum point of the function. Hence, the only solution is at π‘₯=βˆ’2. As a set, this is {βˆ’2}.

Although it was not stated in the previous example, the quadratic function was actually 𝑓(π‘₯)=(π‘₯+2)=π‘₯+4π‘₯+4. Since we know that the root of this function is when π‘₯=βˆ’2, we can create a table of values in this region.

π‘₯βˆ’5βˆ’4βˆ’3βˆ’2βˆ’101
𝑓(π‘₯)9410149

On the graph below, we have plotted these particular points and have marked the single root in red. The π‘₯-axis has also been highlighted in red and we can see that this intersects the function 𝑓(π‘₯) precisely at the point (βˆ’2,0).

This case of a quadratic function only having a single solution is not the most typical one, however. In the following example, we will consider what happens when the parabola crosses the π‘₯-axis at two distinct points.

Example 2: Finding the Solution Set of a Quadratic Equation Graphically

The diagram shows the graph of 𝑦=𝑓(π‘₯). What is the solution set of the equation 𝑓(π‘₯)=0?

Answer

Recall that to find the solutions to 𝑓(π‘₯)=0, we can find the points on the graph of 𝑦=𝑓(π‘₯) where the curve intersects the π‘₯-axis.

We can see that the curve crosses the π‘₯-axis at two points, (βˆ’3,0) and (1,0). This means that the two solutions are βˆ’3 and 1. As a solution set, we can write this as {βˆ’3,1}.

Both examples so far have been of quadratics where the parabola has a β€œu” shape, but they can also have β€œn” shapes. In the next example, we will demonstrate that the method is exactly the same for these instances.

Example 3: Finding the Solution Sets of Quadratic Equations from Their Graphs

The graph shows the curve with equation 𝑦=𝑓(π‘₯). What is the solution set of the equation 𝑓(π‘₯)=0?

Answer

We recall that 𝑓(π‘₯)=0 is satisfied at the points where the graph of 𝑦=𝑓(π‘₯) crosses or touches the π‘₯-axis.

Examining the graph, this happens at the two points (βˆ’2,0), and (2,0). This means that βˆ’2 and 2 are solutions to 𝑓(π‘₯)=0. As a set, this is {βˆ’2,2}.

Let us look at one more example of examining a graph to find solutions, although this time we will consider the case where the graph does not intersect the π‘₯-axis at all.

Example 4: Identifying the Solution Set of a Quadratic given Its Graph

The graph shows the function 𝑓(π‘₯)=π‘₯βˆ’2π‘₯+3. What is the solution set of 𝑓(π‘₯)=0?

Answer

In order to find the solutions to an equation of the form 𝑓(π‘₯)=0, recall that we can identify the points where the graph of 𝑦=𝑓(π‘₯) intersects the π‘₯-axis.

However, if we examine the graph, we can see that there is no point where this occurs. In fact, the minimum point of the graph is at 𝑓(π‘₯)=2. Therefore, there are no solutions to 𝑓(π‘₯)=0. In set form, we can write this as the empty set βˆ….

Although the question does not require us to demonstrate that 𝑓(π‘₯) has no solutions, we will give a quick explanation. Suppose that we tried to find the roots of 𝑓(π‘₯) by completing the square. Then, we can write 𝑓(π‘₯) as follows: 𝑓(π‘₯)=π‘₯βˆ’2π‘₯+3=(π‘₯βˆ’1)βˆ’1+3=(π‘₯βˆ’1)+2.

If we were trying to find the points where 𝑓(π‘₯)=0, then we would be attempting to solve the equation (π‘₯βˆ’1)+2=0.

This is equivalent to requiring that (π‘₯βˆ’1)=βˆ’2, and it is at this point that it is clear why there are no real values of π‘₯ that allow this. If we were to try solving the above equation for π‘₯, then we would first have to eliminate the squared term on the left-hand side by taking the square root of both sides. However, this is not possible given that the right-hand side is negative and we cannot take the square root of a negative number. For this reason, there are no real values of π‘₯ that solve the above equation and hence there are no solutions to the equation 𝑓(π‘₯)=0. Stated slightly differently, there are no roots of 𝑓(π‘₯) and hence the associated solution set 𝑆 is empty.

In the examples above, we were given the graph of different quadratic functions 𝑓(π‘₯), which allowed us to visually identify the points where these graphs crossed the π‘₯-axis. This information allowed us to write these roots of the function in terms of the associated solution set 𝑆, which could have either two distinct elements, one single element, or no elements at all. In the case of there only being a single element, the location of the root would be common with the only vertex of the function.

Even though we were provided with the graphs of the quadratic functions, this information is not necessary if we are only interested in writing the solution set. Helpful information in determining the solution set can be as follows: the precise coordinates of the roots; the sign of the leading coefficients (which determines whether the graph is a β€œu” shape or an β€œn” shape); and the location of the vertex. It is important to note that knowing this information is not the same as knowing each of the coefficients π‘Ž, 𝑏, or 𝑐 for a quadratic 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨. To clarify, just because we know the solution set of a quadratic function, it does not mean that we know what the function is. This is because two or more quadratic functions can have the same solution set, meaning that complete knowledge of the solution set is not enough to know the quadratic function itself (although, it does tell us most of the information that we would require).

As an example, consider the two quadratic functions 𝑓(π‘₯)=π‘₯+3π‘₯βˆ’4,𝑓(π‘₯)=βˆ’2π‘₯βˆ’6π‘₯+8.

These two functions are plotted below, with 𝑓(π‘₯) being plotted in red and 𝑓(π‘₯) plotted in blue. We can see that they both have the solution set 𝑆={βˆ’4,1} due to the roots which are marked at (βˆ’4,0) and (1,0) respectively. Not only are these functions clearly different, but they have a fundamentally different shape, which is due to the different sign on the leading coefficient of each function. Actually, the two functions are related by the formula 𝑓(π‘₯)=βˆ’2𝑓(π‘₯), which explains why these two functions share a solution set, since the roots that populate this set must solve the equation 𝑓(π‘₯)=βˆ’2𝑓(π‘₯)=0. More generally, any two quadratics 𝑓(π‘₯)=π‘˜π‘“(π‘₯) will have an identical solution set for any constant π‘˜β‰ 0. This reiterates the earlier statement that knowledge of the solution set is not enough to fully describe the quadratic function itself.

We can better understand this situation by recalling how the roots of a quadratic correspond to its factors. Assume that there is a quadratic function 𝑓(π‘₯) with the solution set 𝑆={𝑠,𝑠}, where π‘ β‰ π‘ οŠ§οŠ¨. In this case, it is clear that the function 𝑓(π‘₯)=(π‘₯βˆ’π‘ )(π‘₯βˆ’π‘ ) has the given solution set 𝑆, since 𝑓(𝑠)=𝑓(𝑠)=0. It is also the case that any function 𝑔(π‘₯)=π‘˜π‘“(π‘₯) will have the solution set 𝑆 for any π‘˜β‰ 0, since this new function 𝑔(π‘₯) is just a scaling of the original function 𝑓(π‘₯).

If the solution set consists of one element 𝑆={𝑠}, then the only root will be common with the vertex of the function. In this case, the function 𝑓(π‘₯)=(π‘₯βˆ’π‘ ) will have the given solution set, since 𝑓(𝑠)=0 by definition. As with the previous situation, we can multiply the original function 𝑓(π‘₯) by a nonzero constant, meaning that the new function 𝑔(π‘₯)=π‘˜π‘“(π‘₯) will have the same solution set.

We can actually make a more inclusive statement regarding quadratic functions that share the solution set βˆ…. We can say that all quadratic functions that cannot be factored share the same solution set with each other. This means that if a quadratic has no roots, there is very little we can say about it graphically, other than that it cannot intersect the π‘₯-axis. It is worth stating however, that a consequence of this is that the entirety of the curve has to be either above or below the π‘₯-axis. So, if we are given that the vertex is below the π‘₯-axis, then the rest of the curve must be as well, and it is the same above the π‘₯-axis.

In the following examples, we will attempt to answer the questions in a fairly brisk manner, but we will at points reiterate this point above by demonstrating a range of quadratic functions that hold the desired properties.

Example 5: Finding the Solution Set of a Quadratic Function Based on a Description of a Graph

If the graph of the quadratic function 𝑓 cuts the π‘₯-axis at the points (βˆ’3,0) and (βˆ’9,0), what is the solution set of 𝑓(π‘₯)=0 in ℝ?

Answer

To answer this question, we just need to recall that the solutions to a quadratic function are the π‘₯-coordinates of the points where the curve intersects the π‘₯-axis.

Although we do not have the actual graph, we have been told that this happens at (βˆ’3,0) and (βˆ’9,0), so π‘₯=βˆ’3 and π‘₯=βˆ’9 are the roots of the function. As a set, this is {βˆ’9,βˆ’3}.

Note that, in the previous question, we did not have to concern ourselves with the location of the vertex of the curve, although since parabolas are symmetric, we can say that it must be halfway between the two roots (i.e., at π‘₯=βˆ’6).

In the next example, we will demonstrate how in certain cases the vertex will allow us to determine the solutions to a quadratic.

Example 6: Finding the Solution Set of a Quadratic Function given the Vertex

If the point (9,0) is the vertex of the graph of the quadratic function 𝑓, what is the solution set of the equation 𝑓(π‘₯)=0?

Answer

Recall that the solution set of the equation 𝑓(π‘₯)=0 is the set of points where the graph intersects the π‘₯-axis. In this example, we have only been told one point where this intersection occurs, which is at the vertex (9,0) (we know this point intersects the π‘₯-axis since the 𝑦-coordinate is zero).

However, since this point is a vertex, we know that this must be the only point where the graph intersects the π‘₯-axis. This is because the shape of a quadratic graph means the vertex is either the minimum or the maximum point of the curve, and so all other points on the curve must be either above or below the vertex. We demonstrate one example of such a quadratic below.

Thus, the only solution is π‘₯=9, leading to the set {9}.

For our final example, let us see another instance of how the vertex of a quadratic function can tell us information about the solutions.

Example 7: Understanding the Geometry of a Quadratic Function and Using This to Determine Its Solution Set

True or False: If the point (βˆ’2,βˆ’5) is the vertex of the graph of the quadratic function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨ where π‘Ž is a negative number, then the solution set of the equation 𝑓(π‘₯)=0 is βˆ….

Answer

We first recall that the solution set of 𝑓(π‘₯)=0 is βˆ… if the graph of the function does not intersect the π‘₯-axis at any point. We have not been explicitly told whether this is the case, but we do know the location of the vertex. Furthermore, we have been told that π‘Ž is negative, which means that the quadratic equation will have an β€œn” shape as opposed to the β€œu” shape that is found when π‘Ž is positive.

This means that the vertex is the maximum point of the graph, and all other points will be below this point. Since the 𝑦-value of the vertex is βˆ’5, this means 𝑓(π‘₯)β‰€βˆ’5 for all values of π‘₯. We illustrate what this graph might look like below (but note that there are an infinite number of possible graphs).

Thus, it is impossible for 𝑓(π‘₯) to ever be zero, which means that the solution set is indeed empty. So, the answer is β€œTrue.”

Let us finish by considering the key points we have learned during this explainer.

Key Points

  • A quadratic can be graphed using a table of values and this can be used to approximate the points at which the graph intersects the π‘₯-axis. These points are known as the β€œroots” of the quadratic.
  • There are three cases to consider when solving quadratic equations using the graph of 𝑦=𝑓(π‘₯):
    • If the curve crosses the π‘₯-axis twice, 𝑓(π‘₯)=0 has two solutions.
    • If the curve touches the π‘₯-axis at a single point, 𝑓(π‘₯)=0 has one solution.
    • If the curve never intersects the π‘₯-axis, 𝑓(π‘₯)=0 has no solutions.
  • When the solution set only has one element, the graph of 𝑓(π‘₯) touches the π‘₯-axis once at the vertex (either a minimum or maximum point).

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