Joshua F.

asked • 09/03/23

Need help with finding a function

I'm trying to find a function with a Horizontal asymptote of -6, Vertical asymptotes of -1 and 3. It is supposed to have x intercepts of (-3,0) and (1,0). When I was using the asymptotes to find the equation I got (-6x^2)/(x^2-2x-3) but those don't appear to solve it.

Doug C.

Hi Joshua, Check out this graph to see if it helps you understand. desmos.com/calculator/r095wqts57 Post a reply here if you still need some help.
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09/03/23

Doug C.

Here is another graph with the horizontal asymptote included. desmos.com/calculator/wjsfevn25w
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09/03/23

2 Answers By Expert Tutors

By:

Josh F.

tutor
Yes to all except probably Hediye means c = -6, the given y-value of the horizontal asymptote. Btw, it is instructive to leave both numerator and denominator in factored form, as we can then see by inspection how the given information allows us to write the equation of the rational function: f(x) = [-6(x - 1)(x + 3)] / [(x - 3)(x + 1)] The zeros of the numerator generate the x-intercepts of the graph. The zeros of the denominator generate vertical asymptotes. A rational function will have a horizontal denominator different from the x-axis only when the degrees of the numerator and denominator are = , in which case the horizontal asymptote occurs at y = the ratio of the lead coefficients of the numerator and denominator (in this case -6 and 1). From one Joshua F. to another : )
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09/04/23

Hediye G.

tutor
Yes you are right, it should be -6. I remembered HA as 6 and the answer reflects that. Thank you Joshua.
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09/04/23

James S.

tutor
It might be worthwhile to give a brief explanation of BOBO, BOTN, and EATSDC here.
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09/06/23

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