Solve the rational inequality and graph the solution set on a real number line. Express the solution set in interval notation. So if we thought about having the equation equaling Y and we graphed it, we'd realize that we have an X intercept when the numerator is equal to 0. So at -8 zero and 10 zero, we'd have AY intercept when we put zero in for all the XS, and we'd have -80 / 6 or -40 thirds. Vertical asymptotes occur when the denominator equals 0, so X equal -6 and the degree on top is higher. So we have to do polynomial division. Or we could actually do synthetic in this particular case because it's a linear term in the denominator. I'm going to multiply that X + 8 * X - 10 out first and get X ^2 - 2 X -80. Divide it by X + 6. We get X - 8 - 32 / X + 6. Now remember we're talking about what happens when we go out to Infinity and negative Infinity. So that remainder is going to go to 0. So we're really thinking about the line Y equal X - 8. So if we do a rough sketch, we'd have X intercepts at -8 zero and 10 zero. We'd have AY intercept at zero -40 / 3, vertical asymptote at X equal -6, and an oblique asymptote at Y equal X -, 8. So when we graph this, we know when we're at a positive Infinity, we have to be close to that oblique asymptote. When we get to the .10 zero, its multiplicity was one. So if the Y values are on one side positive, the Y values on the other have to be negative. We're going to go to the Y intercept and we're going to continue on to be towards the vertical asymptote. We know we weren't going to go back up because we didn't have an X intercept over there. So now the vertical asymptote, which came from the denominator has a multiplicity of one. So if the Y values on one side are negative, the Y values on the other side are going to be positive. So we're going to come in from the positive direction for the Y values on the other side of the asymptote. We're going to come down to the X intercept -8 zero. We're going to see that its multiplicity, that terms in the numerator is one. So the Y values on one side were positive. the Y values on the other side are going to be negative. And we have to get close to that oblique asymptote as we go out to Infinity. Now the problem actually asks for how to know when it's less than or equal to 0. So if we're talking about less than or equal to 0, we're talking about Y values that are negative. So I want all of these Y values, and I want these Y values, and I want them in terms of the X's. So when I look at this, I can see that my X value has a negative Y value from negative Infinity up until I hit that intercept, which is -8 And because it said less than or equal to, we're going to include the -8. Now there's more than one spot that that's true. So we're going to have a union. It's going to start being true at the opposite side or to the right of that asymptote. So from -6 all the way to the next intercept over here, which is 10, we're going to use bracket at 10, but parenthesis at -6 because we get closer and closer and closer to -6 but we never touch it.