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Graphing a Rational Expression  Problem 3
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Graphing a basic rational expression with transformations. We are now going to look at a problem where we have a negative coefficient and if you think back to how this affects are polynomial graphs we can sort of take the same thing and extract it to this problem.
Let’s take a look at f(x) is equal to x cubed. If you remember this problem we end with a graph that looks something roughly like this. If we throw a negative in front of that, what that did was it took all the y values that were positive, made it negative and all the y values that were negative, and made them positive. Basically just flipped it over the y axis, so we have them something like this.
The basic premise holds shoe for 1 over x as well. Let’s go back over here, we know that the graph of one over x looks something like that, so what ends up happening when we have the negative out in front is the part that’s negative becomes positive and the part that’s positive then becomes negative.
Same exact flip just the different shape of the graph. So whenever you are dealing with any rational expression and you have a negative on the front just flip it over the asymptote it's everything else stays the same, same shape different quadrants.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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