Posted by Soroban on September 29, 2002 at 12:44:51:
In Reply to: Rational functions posted by Jimmu on September 29, 2002 at 11:03:18:
There are some "tricks" to writing these functions.
: Write a rational function whose graphs have the characteristics described in parts a-c. Explain why you selected each problem
: a. A hole at x=-1
f(x) = (2x + 2)/(x + 1) is one of the simplest functions.
f(-1) = 0/0 does not exist.
For any x =/= -1, f(x) = 2, a horizontal line.
You can create your own "function with a hole".
Start with any expression, say, 2x, and multiply by (x + 1)/(x + 1):
f(x) = (2x^2 + 2x)/(x + 1)
The graph will be the line y = 2x with a "hole" at x = -1.
: b. A vertical asymptote at x=-1 and a horizontal asymtote
For a vertical asymptote at x = -1, use (x + 1) in the denominator.
For a horizontal asymptote, make the numerator the same degree, say, 2x + 3.
f(x) = (2x + 3)/(x + 1) has the horizontal asymptote: y = 2.
Since (2x+3)/(x+1) = 2 + 1/(x+1), we can see that as x -> oo,
the fraction -> 0, and f(x) approaches y = 2.
: c. A vertical asymptote at x=-1 and an oblique asymptote
Again, use (x + 1) in the denominator.
This time, make the numerator a higher degree, say x^2 + x + 2:
f(x) = (x^2 + x + 2)/(x + 1)
Since (x^2 + x + 2)/(x + 1) = x + 2/(x+1), we can see that, as x -> oo,
the fraction -> 0, and the function approaches the oblique line y = x.
Hope that helps...
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