Question Video: Identifying the Graph of a Quadratic Equation Mathematics • Third Year of Preparatory School

Video Transcript

Which of the following graphs could represent the quadratic equation of the form 𝑓 of 𝑥 is equal to 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, where 𝑎 is a positive number, given that the solution set to the equation 𝑓 of 𝑥 is equal to zero is the set containing one and two?

In this question, we’re given five possible graphs, and we need to determine which of these five graphs represents a quadratic equation whose leading coefficient is positive and where the solution set to the equation 𝑓 of 𝑥 is equal to zero is the set containing one and two. And we’ll go through all five of the options. However, before we do this, let’s look at the information we’re given. We can start by recalling the graph of any quadratic function has a parabolic shape, and the orientation of this parabola is determined by the sign of its leading coefficient.

If the leading coefficient is negative, we say the parabola opens downwards. However, if the leading coefficient is positive, then we say that the parabola opens upwards. In this case, the sign of the leading coefficient 𝑎 is positive. So we must have a parabola which opens upwards.

Next, we’re told the solution set to the equation 𝑓 of 𝑥 is equal to zero is the set containing one and two. And we can recall the solution set to an equation is the set of all values which satisfy the equation. So because one and two satisfy this equation, 𝑓 evaluated at one must be equal to zero, and 𝑓 evaluated at two must also be equal to zero. And this tells us some information about the graph. If 𝑓 evaluated at one is equal to zero, then the point with coordinates one, zero must lie on our graph. Similarly, if 𝑓 evaluated at two is equal to zero, the point with coordinates two, zero also lies on our graph. This means we have two 𝑥-intercepts for our graph, and we know parabola can have at most two 𝑥-intercepts. So these are all of the 𝑥-intercepts.

And this is enough information to sketch a general idea of what our curve will look like. It’s a parabola which opens upwards and it has two 𝑥-intercepts, one at one and one at two. But we don’t know the exact shape of this curve. For example, we don’t know its 𝑦-intercept. There are infinitely many parabola which open upwards which pass through the points one and two. For example, we could have a parabola which is more narrow, or we could have one which is wider.

However, we can use this information to eliminate some of our options. So let’s start by looking at option (A) as shown. In this case, we do have a parabolic shape. However, we can see that this opens downwards, so its leading coefficient is negative. In our case, our parabola must open upwards because its leading coefficient is positive. So option (A) is not the correct answer.

Let’s now take a look at option (B). We can see this is a parabolic shape and it does open upwards. However, if we look at the coordinates of the 𝑥-intercepts, we can see these are not at one and two. In fact, one of these 𝑥-intercepts is negative. So if we call the quadratic function graphed 𝑔 of 𝑥, then the solution set to the equation of 𝑔 of 𝑥 is not the set containing one and two since one of its elements must be negative. Therefore, option (B) cannot be the correct sketch. Its 𝑥-intercepts are not correct.

We get a similar story if we look at option (C). It is a parabolic shape and it does open upwards. However, we can see that its 𝑥-intercepts are negative. In fact, its 𝑥-intercepts are at negative one and negative two, not one and two. So option (C) is not correct. If we look at option (D), we can see it’s a parabolic shape which opens downwards, which means its leading coefficient is negative. However, we’re told the leading coefficient is positive, so our parabola opens upwards. Option (D) is not correct.

Finally, let’s look at option (E). First, we can see it is a parabolic shape, and we can see it does open upwards, so its leading coefficient is positive. Next, we can take a look at the two 𝑥-intercepts. We can see that these are 𝑥 is equal to one and 𝑥 is equal to two. So if we call this function 𝑔 of 𝑥, then the solution set to the equation 𝑔 of 𝑥 is equal to zero will be the set containing one and two since these are its 𝑥-intercepts. And this agrees with the information we’re given. So the answer is option (E).