Video: Graphing Quadratic Functions

In this video, we will learn how to sketch the graph of a quadratic function that is given in its standard or vertex forms, and to recognise the equation of a quadratic function given its graph.

17:27

Video Transcript

In this video, we will learn how to graph quadratic functions. We will consider those given in standard and vertex form. Quadratic function graphs have many applications. They can represent the paths of moving objects like a bouncing ball or even flight patterns of bees. In business, they can forecast revenue. We could also do something like minimize waste and package sizing. When working with quadratic function graphs, we often deal with variables such as speed, cost, and area, among other things.

Before we look at any specific graphs, let’s remind ourselves of the general form of a quadratic function. When we talk about quadratic functions, the general form would be 𝑓 of π‘₯ equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐, where π‘Ž, 𝑏, and 𝑐 are constants and π‘Ž cannot equal zero. Another form we might see quadratic functions in looks like this. 𝑓 of π‘₯ equals π‘Ž times π‘₯ minus β„Ž squared plus π‘˜, where π‘Ž, β„Ž, and π‘˜ are constants and π‘Ž is not equal to zero. We call this form the vertex form. It’s called the vertex form because when our quadratic function is written in this form, the vertex of this graph will be at the point β„Ž, π‘˜.

If we have this sketch of a quadratic function, the vertex would be found here. The vertex is the maximum in this case because the graph opens downward. And when the quadratic opens upward, the vertex is instead a minimum. We know that when π‘Ž is greater than zero, that is, when π‘Ž is positive, the graph will open upward. And when π‘Ž is less than zero, when π‘Ž is a negative value, the graph will open downward. We noted with the vertex form on the right that when written in this form, the vertex is found at the point β„Ž, π‘˜. When we consider the general form, we’re not able to identify the vertex simply by looking at the general form.

However, we do have a formula to find the vertex if we have a function in general form. From the general form, we take the negative 𝑏-value and divide it by two times the π‘Ž-value. This gives us the π‘₯-coordinate of the vertex. We take that π‘₯-coordinate and we plug it back into our function to solve for the 𝑦-coordinate of the vertex. Before we move on, we should mention another vertex form. Some people in textbooks might use the form 𝑓 of π‘₯ equals π‘Ž times π‘₯ plus 𝑝 squared plus π‘ž, where π‘Ž, 𝑝, and π‘ž are constants and π‘Ž does not equal zero.

This form operates very similarly to the first vertex form we looked at, with one key distinction. If you’re using this form, the vertex is found at negative 𝑝, π‘ž. By looking closely at these two functions, we can see why. In the first form, we’re subtracting β„Ž from π‘₯. And then we have a positive vertex π‘₯-coordinate. In the second form, we are adding that constant to π‘₯, and we’ll therefore need to take the negative value as the π‘₯-coordinate for the vertex. Both forms are equally acceptable as long as you’re using them consistently.

To prepare to graph a quadratic function, we’ll consider four different features that will help us graph these functions. First, we’ll consider the shape of the parabola. Second, we’ll consider the 𝑦-intercept of quadratic functions. Third, we’ll consider the roots or zeros of the function, which are the π‘₯-intercepts of the function. And finally, we’ll consider the vertex or turning point of the function. To help us consider each of these features, we’ll look at the function 𝑓 of π‘₯ equals π‘₯ squared minus two π‘₯ minus eight. To find the shape of the parabola, we’ll want to know is the π‘Ž-value greater than zero or less than zero? That is, is the π‘Ž-value positive or negative?

Positive π‘Ž-values open upward, and negative π‘Ž-values open downward. In our case, the coefficient of the π‘₯ squared term is one. One is greater than zero. Therefore, the shape of this graph will open upward. Next, we want to find the 𝑦-intercept. The 𝑦-intercept will be the place when π‘₯ equals zero. For us, that is zero squared minus two times zero minus eight. 𝑓 of zero equals negative eight, which makes the 𝑦-intercept zero, negative eight. It’s worth pointing out here that when we’re dealing with the general form of a quadratic function, the 𝑦-intercept is always found at the point zero, 𝑐. In our function, 𝑐 is equal to negative eight, which confirms a 𝑦-intercept of zero, negative eight.

What about the roots of a quadratic function? The roots are the place when the 𝑓 of π‘₯ equals zero. To find the roots, we set our 𝑓 of π‘₯ equal to zero and then solve for π‘₯. When the equation is given in general form, we can solve by factoring. We have two π‘₯-terms. We need factors that when multiplied together equal negative eight and when added together equal negative two, which will be negative four and positive two. We then need to take each of these factors and set them equal to zero. We need to know when π‘₯ plus two would be equal to zero and when π‘₯ minus four would be equal to zero. That would be when π‘₯ equals negative two and when π‘₯ equals positive four. This is telling us that this quadratic function graph will cross the π‘₯-axis at the point negative two, zero and the point four, zero.

Our final feature, the vertex, would be given in the form β„Ž, π‘˜ if our original function was written in vertex form. But since we only have the general form, we’ll need to calculate negative 𝑏 over two π‘Ž for the π‘₯-coordinate and then plug that in to solve for the 𝑦-coordinate of this vertex. Negative 𝑏 over two π‘Ž would be the negative of negative two, so positive two over two times one. Positive two over two equals one. This means that the π‘₯-coordinate of our vertex is one. To find the 𝑦-coordinate of the vertex, we plug in one for π‘₯. We get one squared minus two times one minus eight, which equals negative nine. And we can say the vertex of this function is located at one, negative nine. And because we know the shape of this function, we can say that the vertex will be a minimum coordinate.

And with these key features, we’re ready to sketch a graph. We know our 𝑦-intercept is located at zero, negative eight. We have roots at negative two, zero and four, zero. And our vertex, our minimum coordinate, is at one, negative nine. You might also remember that these graphs are symmetric about the axis of symmetry, which is located along the line of the π‘₯-coordinate of the vertex. Thinking about the axis of symmetry gives us a better idea of how to sketch the graph. And we’ll do that now by connecting these points. And following all these features, we’ve graphed a quadratic function. Using these four features, we’re ready to consider some examples.

Write the quadratic equation represented by the graph shown.

In order to write this equation, let’s consider some features of the graph of quadratic functions, shape, 𝑦-intercept, the roots, and the vertex. We also have the general form 𝑓 of π‘₯ equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 or the vertex form 𝑓 of π‘₯ equals π‘Ž times π‘₯ minus β„Ž squared plus π‘˜. Considering the shape, we see that this parabola opens downward. And that means the π‘Ž-value will be less than zero. It will be negative. The 𝑦-intercept is located here at zero, zero, which mean that the roots here will just be one root at zero, zero. And it happens to be the vertex, which is the maximum point of this function.

If we use vertex form, we know that the vertex is found at the point β„Ž, π‘˜. And so we can take this vertex of zero, zero and plug it into that general vertex form. When we simplify that, we find out that the function is π‘Ž times π‘₯ squared. And that means we need to know what π‘Ž is. We know that the π‘Ž-value is negative. But in order to find what it exactly is, we need to consider another point from the graph. We could use one of the other points we know from the graph. For example, we know that the graph crosses the point two, negative four. So we plug in two for π‘₯ and negative four for 𝑓 of π‘₯. And then we have negative four equals four π‘Ž.

From there, we divide both sides of the equation by four. And we see that negative one equals π‘Ž or π‘Ž equals negative one. And we plug that value back in for π‘Ž, which gives us negative one times π‘₯ squared. And we can simplify that to just be negative π‘₯ squared. So we found the quadratic equation represented by the graph shown to be 𝑓 of π‘₯ equals negative π‘₯ squared.

In our next example, we start with the quadratic equation and we need to find its graph.

Which of the following graphs represents the equation 𝑦 equals negative two π‘₯ squared plus nine π‘₯ minus seven?

Our function 𝑓 of π‘₯ equals negative two π‘₯ squared plus nine π‘₯ minus seven is given in the general form π‘Žπ‘₯ squared 𝑏π‘₯ plus 𝑐. But before we do anything with the function, let’s consider each of these graphs. When we look at the graph of (A), it opens upward and has a vertex in the fourth quadrant. The graph of (B) opens downward and has a vertex in the second quadrant. (C) opens downward and has a vertex in the first quadrant. (D) opens downward and has a vertex in the first quadrant. And (E) opens upward and has a vertex in the fourth quadrant.

A good strategy to finding the graph would be to take our equation and find its vertex. When we have an equation in the general form, we find the vertex to be negative 𝑏 over two π‘Ž. That’s the π‘₯-coordinate. So we take negative nine over two times negative two, which is positive 2.25. This means the π‘₯-coordinate of our vertex will be located at 2.25. If we draw the line π‘₯ equals 2.25 on all five of our graphs, we see the in (A) that intersects the vertex and in (C) that intersects the vertex. On that basis, (B), (D), and (E) are not the graphs we’re looking for.

To find the 𝑦-coordinate of the vertex, we could plug in 2.25 into our equation. When we solve for that, we get 3.125. And the vertex for (C) is located at 3.125. We also notice that our π‘Ž-value is negative. And that means our graph must open downward. Option (A) opens upward and has a vertex in the wrong place. And so we can say that option (C) is the correct graph here.

Here’s another example where we’ll use a graph to write a quadratic equation.

Write the quadratic equation represented by the graph shown.

We have the general form for a quadratic equation 𝑓 of π‘₯ equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 and the vertex form π‘Ž times π‘₯ minus β„Ž squared plus π‘˜. To find the equation, we’ll consider some features of the graph. The shape of this graph opens downward. And that means we know that our π‘Ž-value will be less than zero. It will be negative. We have a 𝑦-intercept at the point zero, two. When it comes to the roots, we can’t know with a great deal of accuracy what the roots are. So we’ll leave them there for now. The vertex here is a maximum. And we do know where it’s located, at the point one, three.

Because we know the vertex, we can start with our vertex form. Our vertex is β„Ž, π‘˜. And so we plug in one for β„Ž and three for π‘˜. The only thing we’re missing now is this π‘Ž-variable. We know that it’s less than zero, but we don’t know exactly what it is. To find it, we can plug in another point from the graph that we already know. If we plug in zero for π‘₯ and two for 𝑓 of π‘₯, we’ll be able to solve for our π‘Ž-value. Zero minus one squared is one; one times π‘Ž is π‘Ž, which means π‘Ž plus three equals two. And two minus three is negative one, so we can say that π‘Ž equals negative one.

And we’ll go back and plug that in. Instead of having negative one, we can just write the negative sign and say that 𝑓 of π‘₯ equals negative π‘₯ minus one squared plus three. This is the vertex form of the graph we have. If we wanted to write this in the general form, we could expand this π‘₯ minus one squared, which would give us the negative of π‘₯ squared minus two π‘₯ plus one plus three. We distribute the negative. And when we combine like terms, we have the general form of negative π‘₯ squared plus two π‘₯ plus two. Both of these forms are the quadratic represented by this graph.

In our final example, we’re not looking for a vertex form or a standard form of a quadratic function. Instead, we’re going to explore the factored form.

Write the quadratic equation represented by the graph shown. Give your answer in factored form.

To do this, let’s consider some features of the graph of quadratic functions like shape, 𝑦-intercept, roots, and vertex. The shape of this parabola opens upward. And that tells us that our π‘Ž-value will be positive, greater than zero. We have a 𝑦-intercept located at the point zero, zero. To find the roots, we look for the place where the graph crosses the π‘₯-axis. Here we have a root at zero, zero and at five, zero. The vertex of this graph has been labeled. It’s a minimum coordinate located at two and a half, negative 6.25.

What we wanna do now is think a little bit more about the roots. The roots are sometimes called solutions. They’re the place where our function equals zero. We have solutions when π‘₯ equals zero and when π‘₯ equals five. We can take these solutions and turn them into factors. Because π‘₯ is the same thing as π‘₯ minus zero, π‘₯ equals zero is a factor. And for factor two, we would want to say π‘₯ minus five equals zero. Here are two factors. However, we still don’t know what our π‘Ž-value is because there are many equations that have the factors π‘₯ and π‘₯ minus five. And that means we should have 𝑓 of π‘₯ equals π‘Ž times π‘₯ times π‘₯ minus five.

To solve for π‘Ž, we can plug in coordinates of points we know fall on this graph. We know that when π‘₯ equals 2.5, 𝑓 of π‘₯ equals negative 6.25. Two and a half minus five is negative two and a half. Two and a half times negative two and a half equals negative 6.25. And if we divide both sides of the equation by negative 6.25, we see that π‘Ž equals one. And if π‘Ž equals one, then our factored form is 𝑓 of π‘₯ equals π‘₯ times π‘₯ minus five. We always want to check for this π‘Ž-value because as I’ve just drawn on top of this graph β€” here’s an equation that has the same factors. However, in this case, the π‘Ž would have to be less than zero because the graph opens downward, which means it’s not enough just to know the factors. You need to also check for the π‘Ž-value.

Before we finish, we’ll summarize what we’ve learned. Using the forms of quadratic functions, 𝑓 of π‘₯ equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 and 𝑓 of π‘₯ equals π‘Ž times π‘₯ minus β„Ž squared plus π‘˜, we can identify features of the graph. For the general form 𝑓 of π‘₯ equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐, when π‘Ž is greater than zero, the parabola opens upward. When π‘Ž is less than zero, the parable opens downward. The 𝑦-intercept is located at zero, 𝑐.

To find the π‘₯-coordinate of the vertex, you take negative 𝑏 over two π‘Ž. To find the 𝑦-coordinate of the vertex, you find the π‘₯-coordinate and then plug that back into your function. And for 𝑓 of π‘₯ equals π‘Ž times π‘₯ minus β„Ž squared plus π‘˜, the vertex form. When π‘Ž is greater than zero, the parabola opens upward. When π‘Ž is less than zero, the parabola opens downward. And the vertex is located at the point β„Ž, π‘˜.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.