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Graphs with Holes - Problem 1

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Let's graph a rational function that has a hole in it. Here is our rational function and it looks pretty complicated. Usually the first thing I like to do is to plot intercepts and asymptotes, so my habit is to factor these completely, so we'll factor the numerator and the denominator. And the numerators are quadratic starting with x², so it's going to be x plus or minus something times x plus or minus something.

Now 48 has lots and lots of factors, but we need to find two factors that add up or subtract to 14, 6 and 8 work. So let's try 6 and 8, and if we have a minus both here and here, we'll get a nice +48, -6x minus 8x that's -14x, that works.

In the denominator we can just factor out a -2, and then we're left with -6 plus x, x minus 6. This is what creates a hole in the graph of rational function, a common factor in the numerator and the denominator. Now the function is not defined at x equals 6, but it's not going to have a vertical asymptote there, because this function is actually equivalent to -1/2 x minus 8 for x not equal to 6.

So this is nice easy graph to line, very easy to draw. So what I'm going to do is draw this line and wherever I draw a line, I like to plot the two intercepts, the x intercept and the y intercept. And the x intercept is going to be wherever makes this 0, 8, (8,0) that's my x intercept. The y intercept I get by plugging 0 in, so -1/2 times -8 is 4, (0, 4), those are my two intercepts (8, 0) and (0, 4) and so I'll draw my line.

Only this line is not exactly my rational function, I have to show that x can equal to 6, what does that mean? It means there's a little hole right here when x equals 6. And if you want to find out the exact coordinates of the hole, you can plug 6 into this function right? This function is this expression define of 6, it's the domain restriction that tells us that 6 isn't in the domain. So if I plug 6 in, I find out where the hole is, the hole is at 6 comma 6 minus 8 is -2 times -1/2 is 1, the hole's at (6, 1) and this is our graph of the rational function.

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