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Graphs with Holes - Problem 2
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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.

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