Question

Hello, ok i am in need of some serious help. . I need to build a rational function that has a Horizontal asymptote of Y=2, Two vertical asymptotes where one is psitive and the other is negative, Only one x intercept that is negative, and must ave a positive y intercept.. i have Y=(2x+1)^2/x^2-4.. but it doesn't work pease help :/

Answer 1

ok rational functions

vertical asymptotes are easier, lets start with them:
suppose the rational function h(x) is the polynomial f(x) divided by the polynomial g(x):
\[h(x) = \frac{f(x)}{g(x)}\]

vertical asymptotes occur when g(x) = 0

so a good start would be g(x) = (x+a)(x-b) where a, b>0

because then we get a positive and a negative V. asymptotes

lets choose a = b = 1 for now
\[h(x) = \frac{f(x)}{(x+1)(x-1)}\]

Answer 2

---> in denominator (x-2)^2(x+2)

Answer 3

horizontal asymptotes:

now we want to choose f(x) so that as x tends to plus or minus infinity we get y tending to 2

multiplying out g(x) we get:

\[h(x) = \frac{f(x)}{x^2-1}\]

we need to choose a quadratic f(x) with the coefficient of x^2 as 2
so that as x gets large h(x) gets close to 2

Answer 4

while choosing f(x) we may as well make sure we get our intercepts right

your question says we need a negative repeated x intercept, so we should choose something like

\[f(x) = 2(x+ 5)^2\]

which gets us an x intercept of -5 , while giving us a y intecept of 2(5^2) = 50

Answer 5

so \[h(x) = \frac{2(x+5)^2}{x^2-1}\]

Answer 6

of course there are others, this is just one such function that works

does that all make sense?

Answer 7

tell me if you want me to explain any of it more

Answer 8

ok so i see the one x intercept, but can Horizontal asymptoteintersect with lines/

Answer 9

and is -1, and 1 my vertical asymptotes

Answer 10

ah ok so for horizontal asymptotes we think about what happens as x gets really big
expand the denominator and numerator of h(x) i gave you:
\[h(x) = \frac{2x^2 + 20x +50}{x^2-1} = \frac{2x^2 + 20x +50}{x^2-1} \times \frac{x^{-2}}{x^{-2}} =\frac{2 +\frac{20}{x} + \frac{50}{x^2}}{1 - \frac{1}{x^2}}\]

now as x gets reaaallly big our denominator tends to 1 and our numerator tends to 2

Answer 11

could this one work 2(x+1)^3/(x-)^2(x+2)