Lesson Video: Solving Quadratic Equations Using Graphs Mathematics

In this video, we will learn how to graph a quadratic function of the form 𝑓(π‘₯) = π‘Žπ‘₯Β² + 𝑏π‘₯ + 𝑐, in order to solve the equation 𝑓(π‘₯) = 0.

12:35

Video Transcript

In this video, we’ll learn how to graph a quadratic function of the form 𝑓 of π‘₯ equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 in order to solve the equation 𝑓 of π‘₯ equals zero.

Let’s begin by recalling what we know about the graphs of quadratic functions. They’re the shape of a parabola. And the coefficient of π‘₯ squared tells us the orientation of this parabola. In particular, if the quotient of π‘₯ squared that is π‘Ž is greater than zero, we have that U-shaped parabola, and if π‘Ž is less than zero, we get an N-shaped parabola. In other words, if the coefficient of π‘₯ squared is negative, we have the inverted parabola. In fact, we can also sketch such graphs by calculating the values of their π‘₯- and 𝑦-intercepts. The values of their π‘₯- and 𝑦-intercepts are found by setting 𝑦 equal to zero and π‘₯ equal to zero, respectively, then either solving the resulting equation or simply simplifying.

Specifically for a quadratic function of the form 𝑓 of π‘₯ equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐, if the equation π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero has two distinct solutions, π‘₯ equals π‘₯ one and π‘₯ equals π‘₯ two, then the intercepts are π‘₯ sub one and π‘₯ sub two as shown here. Now this is really important when it comes to using the graphs of such functions to solve equations. Since the π‘₯-intercepts of the graph of a function 𝑦 equals 𝑓 of π‘₯ are found by solving the equation 𝑓 of π‘₯ equals zero, the reverse is true. It follows that the solutions to the equation 𝑓 of π‘₯ equals zero can be found by locating the π‘₯-intercepts of the graph.

Now, of course, in the case of quadratic graphes specifically, it’s a little more nuanced than this. We’ve just seen that if the graph of a quadratic function has two distinct π‘₯-intercepts π‘₯ one and π‘₯ two, then the equation 𝑓 of π‘₯ equals zero has two distinct solutions. But there will be occasions where our equation will have one solution, sometimes called a repeated root, or no solutions at all. These can be quickly identified again by looking at the graph of the function. A repeated root will occur when the π‘₯-axis is a tangent to the graph. In other words, the graph touches the π‘₯-axis exactly once.

On these occasions, in fact, the single solution to the equation 𝑓 of π‘₯ equals zero corresponds in fact to the location of the vertex of the graph. Now, if the graph does not intersect the π‘₯-axis at all, then the equation 𝑓 of π‘₯ equals zero has no real roots. This could look like this, in other words, the function has purely positive outputs, or like this, where we have purely negative outputs. So with all of this in mind, we’re going to demonstrate how to use a quadratic graph to solve a quadratic equation.

The diagram shows the graph of 𝑦 equals 𝑓 of π‘₯. What is the solution set of the equation 𝑓 of π‘₯ equals zero?

Remember, given the graph of a quadratic equation 𝑓 of π‘₯ equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐, the solutions to the quadratic equation 𝑓 of π‘₯ equals zero correspond to the values of the π‘₯-intercepts. So all we need to do is identify the location of the π‘₯-intercepts on our graph. Locating these, we see that there is, in fact, only one π‘₯-intercept. It’s negative two. So there’s actually only one solution to the equation 𝑓 of π‘₯ equals zero. That’s π‘₯ equals negative two. Now we have in fact been asked to give the solution set to the equation 𝑓 of π‘₯ equals zero. So in set notation, it’s the set containing the single element negative two.

In our next example, we’ll see how we can perform the exact same process if the graph of the function is an inverted parabola, in other words, if the coefficient of π‘₯ squared is negative.

The diagram shows the curve with equation 𝑦 equals 𝑓 of π‘₯. What is the solution set of the equation 𝑓 of π‘₯ equals zero.

Remember, if we’re given the graph of a function 𝑓 of π‘₯, the solutions to the equation 𝑓 of π‘₯ equals zero correspond to the values of the π‘₯-intercepts. And whilst we’re using this process to find solutions to quadratic equations, this holds to equations of any function 𝑓 of π‘₯ equals zero.

So all we need to do is find the location of the π‘₯-intercepts on our graph. We have two. There’s one here and there’s one here. These are the points of which our graph passes through the π‘₯-axis. Since this occurs at negative two and two, we can say that the solutions to the equation 𝑓 of π‘₯ equals zero are π‘₯ equals negative two and π‘₯ equals two. Now in fact, we were asked to give the solution set. So we use set notation as shown. The solution set to the equation 𝑓 of π‘₯ equals zero is the set containing the elements negative two and two.

We’ll consider one further example of this form.

The graph shows the function 𝑓 of π‘₯ equals π‘₯ squared minus two π‘₯ plus three. What is the solution set of 𝑓 of π‘₯ equals zero.

Given some graph of a function 𝑦 equals 𝑓 of π‘₯, the solutions to the equation 𝑓 of π‘₯ equals zero correspond to the values of the π‘₯-intercepts if they exist. If there are no π‘₯-intercepts, it follows that there are no solutions to the equation 𝑓 of π‘₯ equals zero. Now in this case, we’ve been given the graph of the function 𝑓 of π‘₯ equals π‘₯ squared minus two π‘₯ plus three and asked to solve 𝑓 of π‘₯ equals zero or, in other words, π‘₯ squared minus two π‘₯ plus three equals zero.

But if we look carefully, we see that there are no locations where the graph, which is the green plot here, intersects the π‘₯-axis. And so, there are no real solutions to the equation 𝑓 of π‘₯ equals zero. In order to use set notation, we need to find a way to show that there are no values inside the set. So we use the notation shown. It’s the null set or empty set.

In the examples we’ve been given so far, we’ve been given the graph of some function 𝑓 of π‘₯. And that has allowed us to identify the locations at which this graph crosses the π‘₯-axis. This information then allowed us to write the roots of the function, in other words the solutions, in terms of the associated solution set that could have either two distinct elements, one single element, or no elements at all. And in the case if there only being a single element, the location of the root is common with the only vertex of the function. Now in fact, we won’t always be given the graph of the quadratic function. And it won’t always be obvious how to find exact solutions. In our next example, we’ll plot the graph of a quadratic function and use this to estimate the solutions to the equation.

Plot the graph of the quadratic function 𝑓 of π‘₯ equals two π‘₯ squared plus three π‘₯ minus one over the interval π‘₯ equals negative three and π‘₯ equals one using integer values of π‘₯. Using this graph, estimate the solutions to the equation 𝑓 of π‘₯ equals zero. Give your answers to the nearest whole number.

To plot the graph of a function, we begin by drawing a table of values. Now, since we’re plotting it over the interval π‘₯ equals negative three and π‘₯ equals one and using integer values, we’re going to use π‘₯ equals negative three, negative two, negative one, zero, and one. In order to find the corresponding output, the value of 𝑓 of π‘₯, we substitute each value of π‘₯ into the function two π‘₯ squared plus three π‘₯ minus one. This creates a set of ordered pairs that satisfy our function.

Let’s begin by calculating the value of the function when π‘₯ equals negative three, in other words 𝑓 of negative three. We replace each instance of π‘₯ with negative three, and we get two times negative three squared plus three times negative three minus one. The order of operations then tells us to calculate the exponent in this problem first. Negative three squared is nine. Then we calculate two times nine, which is 18, and three times negative three, which is negative nine. So this becomes 18 minus nine minus one, and that is equal to eight. So 𝑓 of negative three is eight.

We’ll now repeat this process with π‘₯ equals negative two. Our expression is two times negative two squared. Of course, negative two squared is four. Then, we have three times negative two minus one. That becomes eight minus six minus one, which is one. Next, π‘₯ equals negative one. It’s two times negative one squared plus three times negative one minus one. And since negative one squared is one, this becomes two minus three minus one, which is negative two. In a similar way, when π‘₯ is equal to zero, the value of the function is negative one. And when π‘₯ is equal to one, 𝑓 of π‘₯ is equal to four.

And so we have our five ordered pairs. We have a negative three, eight; negative two, one; negative one, negative two; zero, negative one; and one, four. We plot these in turn. Negative three, eight is here. Negative two, one is here. We then have negative one, negative two; zero, negative one; and one, four.

Don’t forget, these outputs don’t increase linearly. So we must join them with a smooth curve instead of a straight line. And so, we have the graph of our function. Remember, we’re trying to use this to find solutions to the equation 𝑓 of π‘₯ equals zero. Now, if these solutions exist, they correspond to the values of the π‘₯-intercepts. These appear to be approximately located at π‘₯ equals negative 1.8 and π‘₯ equals 0.2. Correct to the nearest whole number, an estimate for solutions to the equation 𝑓 of π‘₯ equals zero are π‘₯ equals negative two and π‘₯ equals zero.

Now in fact, we don’t need to be given or drawn the graph to find solutions to 𝑓 of π‘₯ equals zero. We know that the solutions correspond to the values of the π‘₯-intercepts, sometimes equivalently called the zeros of the function. Hence, given these values or coordinates of the π‘₯-intercepts, we can identify a solution set of the equation 𝑓 of π‘₯ equals zero. In our next example, we’ll demonstrate what that looks like.

If the graph of the quadratic function 𝑓 cuts the π‘₯-axis at the points negative three, zero and negative nine, zero, what is the solution set of 𝑓 of π‘₯ equals zero in the set of real numbers?

Now remember, if we’re given the graph of a function, we can find the solutions to 𝑓 of π‘₯ equals zero by locating the π‘₯-intercepts or zeros of the function. Now, in this case, we’re not actually given a graph, but we are told the coordinates at which the function cuts the π‘₯-axis. It’s negative three, zero and negative nine, zero. Since the first number in each ordered pair corresponds to the value of π‘₯ here, we can say that the solutions to the equation 𝑓 of π‘₯ equals zero are π‘₯ equals negative three and π‘₯ equals negative nine. Using set notation, the solution set of 𝑓 of π‘₯ equals zero in the set of real numbers is the set containing the elements negative three and negative nine.

We’ve now demonstrated in a variety of ways how to find solutions to an equation 𝑓 of π‘₯ equals zero when given the graph of the function 𝑓 of π‘₯. Let’s recap the key points from this lesson.

We saw that the solutions to the equation 𝑓 of π‘₯ equals zero, if they exist, can be found by locating the values of the π‘₯-intercepts of the graph of 𝑦 equals 𝑓 of π‘₯. We also saw that if the π‘₯-axis is a tangent to the curve, then we have a repeated root. It’s just one root. And this point is also in fact the vertex of the graph. Finally, we also saw that we can plot quadratic graphs using a table of values. And we can then use this graph to approximate the points at which the graph crosses the π‘₯-axis and hence the solution or the roots of that quadratic equation.

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