Video: Determining Where the Graph of a Quadratic Equation in Factored Form Crosses the π‘₯-Axis

At which values of π‘₯ does the graph of the equation 𝑦 = 5(π‘₯ βˆ’ 1)(π‘₯ + 7) cross the π‘₯-axis?

02:12

Video Transcript

At which values of π‘₯ does the graph of the equation 𝑦 equals five times π‘₯ minus one times π‘₯ plus seven cross the π‘₯-axis?

We begin by recalling that the π‘₯-axis has an equation. We say it’s the line 𝑦 equals zero. And this means we can find the points where our graph crosses the π‘₯-axis by setting 𝑦 equal to zero and solving for π‘₯. This is sometimes called finding the roots of the equation. In this case, we say zero is equal to five times π‘₯ minus one times π‘₯ plus seven. And how do we solve for π‘₯? Well, the first thing you’re going to do is look to get rid of this five. The five is multiplying the expression π‘₯ minus one times π‘₯ plus seven. So, we’re going to divide both sides by five.

When we do, we find that zero is equal to π‘₯ minus one times π‘₯ plus seven. Now, the expressions π‘₯ minus one and π‘₯ plus seven are multiplying one another, and when they do, we get zero. And so, for the expression π‘₯ minus one times π‘₯ plus seven to be equal to zero, that means either π‘₯ minus one itself must be equal zero or π‘₯ plus seven must be equal to zero. We’ll solve the equation π‘₯ minus one equals zero for π‘₯ by adding one to both sides. So, π‘₯ is equal to one.

And we’ll solve the equation π‘₯ plus seven equals zero by subtracting seven from both sides to give us π‘₯ equals negative seven. And so, we have two solutions for π‘₯. For the equation zero equals five times π‘₯ minus one times π‘₯ plus seven, π‘₯ can either be equal to one or it can be equal to negative seven.

Now, of course, before assuming that these are the values we’re interested in, we’re going to check what we’ve done is correct. Let’s take the solution π‘₯ equals one. We substitute it into the expression five times π‘₯ minus one times π‘₯ plus seven. So, we get five times one minus one times one plus seven. That’s five times zero times eight, which is indeed zero, as we expected.

We’ll repeat this for the other value of π‘₯, π‘₯ equals negative seven. That’s five times negative seven minus one times negative seven plus seven. That’s five times negative eight times zero, which is again zero as required. And so, by solving the equation five times π‘₯ minus one times π‘₯ plus seven equals zero, we found the values of π‘₯ for which the graph of the equation 𝑦 equals five times π‘₯ minus one times π‘₯ plus seven crosses the π‘₯-axis to be π‘₯ equals one and π‘₯ equals negative seven.

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