vertical asymptotes: 1/((x+2)(x-7))
root at x = -5:
(x+5)/((x+2)(x-7))
horizontal asymptote. we need the degree to be equal, so try
7x(x+5)
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2(x+2)(x-7)
The problem here is that now we have another x-intercept at (0,0)
So, let's work with
7(x+5)^2
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2(x+2)(x-7)
That gives us an intercept, but it does not cross the x-axis.
We could go with something like
7(x^2+1)(x+5)
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2(x+2)(x-7)^2
That gives us the same degree top and bottom, a single crossing at -5, and still the two vertical asymptotes.
Find a rational function that satisfies the given conditions.
Vertical asymptotes x=-2,x=7
Horizontal asymptote y=7/2
x-intercept (−5, 0)
1 answer
1 answer