 So let's take a detailed look at finding the graph of a rational function. So for the graph of y equals f of x, where f of x is x minus 2 over x plus 1, let's find the x and y intercepts, the vertical and horizontal asymptotes, and then sketch the graph. So remember, the x-intercept will have coordinates h, 0. The x-intercept is the point on the graph where y is equal to 0. So we have y equals f of x equals means replaceable. So f of x is x minus 2 over x plus 1 and y is equal to 0. And that gives us an equation. To solve this equation, we can multiply both sides by the denominator x plus 1. We'll simplify and solve. And remember, the x-intercept is a point and both coordinates must be given. So our point has x equal to 2, y equal to 0, and the x-intercept is at 2, 0. We can also find the y-intercept. The y-intercept will have coordinates 0, k, and so the y-intercept is the point on the graph where x is equal to 0. So we're graphing y equals f of x. f of x is x minus 2 over x plus 1. Equals means replaceable, so we'll replace x with 0. We'll do a little arithmetic and find our y value. And remember, the y-intercept is a point, so both coordinates must be given. We have x equals 0, y equals negative 2, and the y-intercept is at 0, negative 2. Now, since this is a rational function, the denominator can't be equal to 0. We require x plus 1 is not equal to 0. And to solve the inequality, we'll start by solving the corresponding equality, x plus 1 equal to 0, we'll solve, and since x equals negative 1 would make the denominator 0, we forbid x to be negative 1. So we require x not equal to negative 1, and we might note the following. As x gets close to negative 1, but stays slightly more, x plus 1 goes to 0, but slightly more, and x minus 2 goes to negative 3. And so x minus 2 over x plus 1 will go towards a fraction where our numerator is close to negative 3, and our denominator is a small positive number. And so the value will be negative and large, and so we say that x minus 2 over x plus 1 goes to negative infinity. And if it's not written down, it didn't happen. We write down as x goes to negative 1, but stays a little bit more, f of x goes towards minus infinity. On the other side, as x gets close to negative 1, but stays slightly less, x plus 1 goes to 0, but stays a little bit less than 0, and x minus 2 goes towards negative 3. So x minus 2 over x plus 1 will go towards a fraction where our numerator is close to negative 3, and our denominator is a small negative number. And so the value will be positive and large, and so x minus 2 over x plus 1 is going to go towards positive infinity. Again, if it's not written down, it didn't happen. And since as x goes to negative 1 either slightly more or slightly less, our function values go to plus or minus infinity. This means that x equals negative 1 is going to be a vertical asymptote. So to find the horizontal asymptote, we rewrite the rational expression by dividing the numerator and denominator by the highest power in the denominator. So that's x, so we'll multiply numerator and denominator by 1 over x. We'll do a little algebra, and we see what happens as x goes to positive infinity. 2 over x goes to 0, which means that 1 minus 2 over x is going to go to 1. 1 over x goes to 0, 1 plus 1 over x goes to 1, and so in our rational expression, the numerator is going to go to 1, the denominator is going to go to 1, and so the rational expression goes to 1 over 1, otherwise known as 1. And if it's not written down, it didn't happen. As x goes to infinity, our rational function goes to 1. But since we're on the graph of y equals f of x, that means y goes to 1. Similarly, as x goes to minus infinity, 2 over x goes to 0, 1 minus 2 over x goes to 1, 1 over x goes to 0, and 1 plus 1 over x is going to go to 1. So as x goes to infinity, our numerator goes to 1, our denominator goes to 1, and our rational expression goes to 1 over 1, or 1, and if it's not written down, it didn't happen. As x goes to minus infinity, y goes to 1, and so that says that y equals 1 is going to be a horizontal asymptote. And now let's incorporate this information into a graph. We know where the x-intercept is. We know where the y-intercept is. We require x cannot be equal to negative 1, so we'll place a little electric fence line there. If our x values are close to negative 1, but a little bit more, our function values, our y values, are negative and large. So this point way down here. If our x values are close to negative 1, but a little bit less, our function values, our y values, are positive large. So that suggests points up here. As x goes to infinity, if we go to the right, our y values get close to 1. As x goes to minus infinity, if we go to the left, our y values get close to 1. And so altogether our graph smoothed out is going to look something like this.