# 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 π¦ = 5(π₯ β 1)(π₯ + 7) cross the π₯-axis?

02:12

### Video Transcript

At which values of π₯ does the graph of the equation π¦ 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 itβs the line π¦ equals zero. And this means we can find the points where our graph crosses the π₯-axis by setting π¦ 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 youβ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, weβ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. Weβll solve the equation π₯ minus one equals zero for π₯ by adding one to both sides. So, π₯ is equal to one.

And weβ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 weβre interested in, weβre going to check what weβve done is correct. Letβ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. Thatβs five times zero times eight, which is indeed zero, as we expected.

Weβll repeat this for the other value of π₯, π₯ equals negative seven. Thatβs five times negative seven minus one times negative seven plus seven. Thatβ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 π¦ equals five times π₯ minus one times π₯ plus seven crosses the π₯-axis to be π₯ equals one and π₯ equals negative seven.