A certain rational function f(x) contains quadratic functions in both its numerator and denominator. Aside from that, we also know the following things about f:

• f has a vertical asymptote at x = 5
• f has a single x-intercept of x = 2
• f is removably discontinuous at x = 1, since

Based on these delightful clues, evaluate:

(a) f(0)

(b)

Difficulty:

Solution

If f has a vertical asymptote at x = 5, then the factor must appear in the denominator but not the numerator. Why is this? When we plug x = 5 into our function f, we want to get a 0 in the denominator but not the numerator; mathematically, this means an asymptote must appear on the graph for that value of x. On the other hand, the term must appear in the numerator, because substituting x = 2 into f must result in a 0 in the numerator but not the denominator.

That's worth reviewing for a moment. Substituting real number values into a function and getting 0 as a result is important. If you get 0 in the bottom of the fraction but not the top, the result is a vertical asymptote and a nonremovable discontinuity. However, if you get 0 in the top but not the bottom of the fraction, you've got yourself an x-intercept there, mister man.

What happens if you get 0 in both the numerator and the denominator? Good question. It will cause the function to be discontinuous, but a limit will usually exist, so the function is actually removably discontinuous there. Notice that our problem says that f will be removably discontinuous when x = 1 (and even goes so far so to tell you what the limit will be). Therefore, because of what I just got fnished telling you, substituting x = 1 into f has to give us 0 in both the numerator and denominator. Thus, both parts of the fraction contain the factor . Let's see what we've got so far:

The problem tells us that the numerator and denominator are both quadratic functions, and we'll definitely get quadratic functions if we multiply the top and bottom out. So, our answer must look a lot like f. In fact, it can only differ from f by the value of a constant. Why? Because if I multiply f by a constant, it might change the value of its outputs, but it will not change the x-intercept or the locations of discontinuity. Since I don't know what that constant is, I will call it A. Therefore, f must be .

I'd feel better if I could figure out what the heck A was, wouldn't you? Luckily, we can, because we know that . Use this info to find A:

Finally we can figure out that . Now, we can answer those questions I asked, dadgum it:

(a)

(b) Since the numerator and denominator are the same degree, the limit at infinity of f must be A, which we already determined to be .