Help me with creating a rational function with vertical asymptotes x=±1 and oblique asymptote of y=2x−3 and a y-intercept of 4.

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vazen2bl · Accepted Answer

First, you want your function to have vertical asymptotes at x=1,-1. So we really want  1(1−x)(1+x)    (1) Now, as x→∞; this has a &quot;oblique&quot; asymptote of 0, so we can just add the asymptote we want:  1(1−x)(1+x)+2x−3  Now, lets

jorgegalar0xk · Accepted Answer

For one family of rational functions f(x) having vertical asymptotes only at x=±1, consider those that can be written in the formf(x)=p1(x)+p2(x)(x+1)m+p3(x)(x−1)nfor some positive integers m,n, and some polynomials p1,p2,p3, with p2,p3 in particular being non-0 polynomials of respective degrees&amp;lt;m,&amp;lt;n. Such a function has a slant asymptote if and only if P1(x) is the line toward which the function tends asymptotically.Hence, any rational functions f(x) in our family having y=2x-3 for a slant asymptotes can be written in the formf(x)=2x−3+p2(x)(x+1)m+p3(x)(x−1)n,with m,n,p2,p3, as described above.We&#039;re just looking for one such example, so to make it simpler on ourselves, we may as well assume m=n=1, so that p2,p3, must be constant polynomials, and our function will bef(x)=2x−3+Ax+1+Bx−1for some non-zero constants A,B. All that remains is to choose non-zero constants A,B so that f(0)=4 (i.e.: f(x) has y-intercept 4).

Help me with creating a rational function with vertical asymptotes

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