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anonymous
 5 years ago
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 ...
anonymous
 5 years ago
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 ...

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0vertical at x = 3 means denominator should have a factor of \[x3\] in it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0horizontal 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so maybe \[f(x)=\frac{2x}{x3}\] would work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0unfortunately it does not, because \[f(1)=\frac{2}{2}=1\] and you want \[f(1)=0\] so maybe try \[f(x)=\frac{2x2}{x3}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now at least \[f(1)=\frac{22}{13}=0\] how about \[f(\frac{1}{2})\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope that doesn't work, i am a moron numerator should be \[(2x+1)(x1)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that will give you the correct zeros,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so now maybe \[f(x)=\frac{(2x+1)(x1)}{(x3)^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that 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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