Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Graphing a Rational Expression - Problem 2

Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Dealing with the basic rational expression graph and some basic transformations. So for this example what I want to look at is 1 over x plus 3 and before I do that I want to take a look at our basic graph for 1 over x.

So we know the basic shape of 1 over x and we know that we have a horizontal asymptote at 0 and a vertical asymptote at 0 as well. So the reason we have a vertical asymptote at 0 is because we cannot plug 0 into this equation.

When we plug numbers in very close to 0 it's either they are going to go up to infinity or down to negative infinity. We plug in 1 over 100 1 over 100 turns into 100 so we get really close to our y axis we have a huge number.

So that’s why we have this vertical asymptote at 0. Going back over to this problem here what number can we not plug? We can’t plug in -3. So for this case what actually is happening to our problem is our graph is getting moved backwards three units.

We can’t plug in -3 so our vertical asymptote is now at -3. We’ve moved everything back you move a horizontal line backwards it stays in the same place, so what we really end up with is a graph that has the same shape as before everything moved back three, giving us a shape something like that. If this was a x minus we would’ve moved forward same idea what value of x can we not put in that’s where our vertical asymptote is going to be.

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