Mathematics OpenStudy (anonymous):

Create a function that has a graph with features that include -vertical retricemptote at y axis x=3 -horizontal retricemptote at y=2 -x intercept of (-1/2, 0) & (1,0) QUESTION IS REGAURDING RATIONAL FUNCTIONS/QUADRATIC FUNCTIONS ... OpenStudy (anonymous):

asymptotes OpenStudy (anonymous):

lol OpenStudy (anonymous):

vertical at x = 3 means denominator should have a factor of $x-3$ in it OpenStudy (anonymous):

horizontal asymptote at $y=2$ means the degree of the numerator and denominator must be the same, and the ratio of leading coefficients has to be 2 OpenStudy (anonymous):

so maybe $f(x)=\frac{2x}{x-3}$ would work OpenStudy (anonymous):

unfortunately it does not, because $f(1)=\frac{2}{-2}=-1$ and you want $f(1)=0$ so maybe try $f(x)=\frac{2x-2}{x-3}$ OpenStudy (anonymous):

now at least $f(1)=\frac{2-2}{1-3}=0$ how about $f(-\frac{1}{2})$? OpenStudy (anonymous):

nope that doesn't work, i am a moron numerator should be $(2x+1)(x-1)$ OpenStudy (anonymous):

that will give you the correct zeros, OpenStudy (anonymous):

so now maybe $f(x)=\frac{(2x+1)(x-1)}{(x-3)^2}$ OpenStudy (anonymous):

that one will have the correct zeros, and also the correct horizontal and vertical asymptote. i was forgetting that if you have two zeros you need a polynomial of degree 2 !

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