Unit
Polynomial and Rational Functions
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
We're graphing rational functions with holes, here is another example; y equals x³ plus x² minus 6x over 6x. Now the first thing I like to do is factor these expressions so I know where the x intercepts are, and this numerator has a common factor of x. So I'll pull that out, x² plus x minus 6 is left over and the denominator 6x. Now this is a quadratic, it looks factorable so let me factor that. I'll have x plus or minus something, x plus or minus something to give me x². And to give me 6, a 2 and a 3 would work because 3 minus 2 would give me 1. So plus 3 minus 2, that gives me -6, 3x minus 2x is x, perfect.
And now when I have the common factor of x, I can cancel so long as I remember that this equals 1/6 x plus 3, x minus 2 for x not equals to 0. I've gotten rid of the factor in the denominator that makes this function undefined at 0, but I have to record that the domain is still x not equal to 0.
Now this function looks like it's a parabola. This is a quadratic function with a little hole in and since it's factored we know where the x intercepts are. The x intercepts are at -3,0 and 2,0, so let's plot those. Here is -3,0, here is 2,0. And if we plot one or two other points, we'll have enough to draw the graph. Now the hole is going to be at x equals 0, and even though this is a hole, this expression is defined for 0. So I can find the y coordinate of the 0 by plugging in. So the hole is at 0, something, so let me plug zero and I get 3 times -2, -6 times 1/6, -1, that's where our hole is.
So I'll put a little open circle at 0,-1 that's my hole. And then let's try -1, at x equals -1, what is y? Well it's 1/6 of -1 plus 3,2, -1 minus 2, -3, I'm sorry this should be 1/6 and that's 1/6 of -6, -1. At -1 we have a y value of -1, so here is another -1. It looks like our parabola is going to look something like this. It's got a little hole at x equals 0 that's our graph.