Video: Identifying and Writing Quadratic Functions from a Graph

For the function 𝑓(π‘₯) = π‘₯Β² βˆ’ 4π‘₯ + 3, answer the following questions. Find by factoring, the zeros of the function. Identify the graph of 𝑓.

05:20

Video Transcript

For the function 𝑓 of π‘₯ equals π‘₯ squared minus four π‘₯ plus three, answer the following questions. Firstly, find by factoring the zeroes of the function. Secondly, identify the graph of 𝑓.

There are also two further parts to this question. So firstly, we’re asked to find the zeros of this function. And the method we’re told to use is factoring. We therefore need to write our quadratic as the product of two linear factors. As the coefficient of π‘₯ squared is one, we know that the first term in each of our parentheses will be π‘₯. We’re then looking for two numbers whose sum is the coefficient of π‘₯, that’s negative four, and whose product is the constant term, that’s positive three.

Well, the two numbers that fit both of those criteria are negative one and negative three. Negative one plus negative three is negative four, and negative one multiplied by negative three is positive three. So our quadratic factors as π‘₯ minus one multiplied by π‘₯ minus three, which we can of course confirm by redistributing the parentheses if we wish.

We need to use this factored form to determine the zeroes of the function, which we recall are the π‘₯-values such that 𝑓 of π‘₯ equals zero. If we set this factored form equal to zero, we then recall that for the product of two things to be zero, at least one of them must themselves be zero. So we can take each factor in turn and set it equal to zero, giving two simple linear equations. The first can be solved by adding one to each side to give π‘₯ equals one, and the second can be solved by adding three to each side to give π‘₯ equals three. The roots or zeros of this function then are the values one and three.

Now, in the second part of the question, we’re asked to identify the graph of our function 𝑓. And we can see that we’ve been given three possibilities: a blue one, a red one, and a green one. Now, we’ve just found that our graph has zeros at one and three. And remember, these zeros are the values of π‘₯ at which the graph crosses the π‘₯-axis. So if our graph crosses the π‘₯-axis at one and three, we can see from the figure that this leaves just the red and green graphs. The blue graph crosses the π‘₯-axis or has zeros at values of negative one and negative three.

Now, we just need to decide between the red and green graphs, which we see are mirror images of each other. One is an upward-curving parabola, and the other is a downward-curving parabola. We recall that the type of parabola we have will be determined by the value of π‘Ž. That’s the coefficient of π‘₯ squared. In our function, the coefficient of π‘₯ squared is one. It’s a positive value, which means the parabola will curve upwards. That means then that the graph of our function 𝑓 must be the red graph. It has the correct zeros and the correct shape. We can also see that the 𝑦-intercept of this graph is three, which is indeed the constant term in our function 𝑓 of π‘₯.

The remaining two parts of the question, which I didn’t write down initially because they give the game away for the previous part are. Write the equation for 𝑔, the function that describes the blue graph. And write the equation for β„Ž, the function that describes the green graph.

Let’s look at this blue graph first of all then. We already said that it has zeros at negative one and negative three. This means that in its factored form, it has factors of π‘₯ plus one and π‘₯ plus three. But there could also be a factor of π‘Ž that we multiply by. To determine whether the value of π‘Ž is one or something else, we consider the 𝑦-intercept of the graph, which we can see is the same as the 𝑦-intercept of the red graph. It’s three. When we multiply these two factors together, the constant term will be one multiplied by three, which is indeed three. And so this tells us that the value of π‘Ž is simply one. Our function 𝑔 in its factored form then is π‘₯ plus one multiplied by π‘₯ plus three. If we distribute the parentheses, we have 𝑔 of π‘₯ equals π‘₯ squared plus four π‘₯ plus three.

For the green graph, it has the same zeros as our function 𝑓. So it can be written as π‘Ž multiplied by π‘₯ minus one multiplied by π‘₯ minus three. And again, we need to determine whether the value of π‘Ž is one or something else. Well, the 𝑦-intercept for the green graph is negative three. If we multiply together negative one and negative three, we get a value of positive three. And so in order to ensure the 𝑦-intercept, the constant term in the expanded form of β„Ž of π‘₯, is negative three, we need the value of π‘Ž to be negative one.

The equation β„Ž of π‘₯ then is negative π‘₯ minus one multiplied by π‘₯ minus three. In fact, it is the complete negative of our function 𝑓 of π‘₯, which we can also see because they are reflections of one another in the π‘₯-axis. We can write the equation β„Ž of π‘₯ then as the complete negative of our function 𝑓 of π‘₯. β„Ž of π‘₯ is equal to negative π‘₯ squared minus four π‘₯ plus three.

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