SOLUTION: Graph the rational function: r(x)=(x-4)(x+1)/(x-1)(x+4)

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Question 1113853: Graph the rational function:
r(x)=(x-4)(x+1)/(x-1)(x+4)

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Graph the rational function:
r%28x%29=%28%28x-4%29%28x%2B1%29%29%2F%28%28x-1%29%28x%2B4%29%29

(1) zeros: The function value is 0 whenever the numerator is 0 -- when x=4 or x=-1.
(2) vertical asymptotes: The function is undefined and so has a vertical asymptote whenever the denominator is 0 -- when x=1 or x=-4.
(3) horizontal asymptote: As x gets very large (positive or negative) the constants become insignificant; the function value approaches x^2/x^2 = 1; the horizontal asymptote is y=1.

(4) intervals where the function value is positive or negative:

The x values -4, -1, 1, and 4 where the numerator or denominator is zero break the x axis into intervals; we need to determine whether the function value is positive or negative in each interval.

Most sources I have seen will tell you to use a test value in each interval. I find that process too repetitive; you can accomplish the same thing with less work if you simply imagine "walking" along the x axis and see what happens to the sign of the function value each time you cross from one of the intervals to the next.

For this problem, we can see that all 4 factors in the numerator and denominator are negative for large negative values of x; that means the function value is positive from negative infinity to x=-4.
Then when we pass x=-4, the sign of one factor changes; that changes the sign of the function value. So when we pass x=-4, the function value becomes negative.
Similarly, the function value changes from negative to positive when we pass x=-1; from positive to negative when we pass x=1, and finally from negative to positive when we pass x=4.

(5) Last we should find the values of x, if any, where the graph of the function crosses the horizontal asymptote. To do that, since the horizontal asymptote is y=1, we need to find all the solutions to
%28%28x-4%29%28x%2B1%29%29%2F%28%28x-1%29%28x%2B4%29%29=1
%28x-4%29%28x%2B1%29+=+%28x-1%29%28x%2B4%29
x%5E2-3x-4+=+x%5E2%2B3x-4
6x=0
x=0

The function value crosses the horizontal asymptote at x=0, and only there.

With all of the above, the basic behavior of the function is determined.

Here is a graph...: