Arthur D. answered • 09/07/15

Effective Mathematics Tutor

[(x-1)(x-3)]/[2x(x+1)]

you are correct

the vertical asymptotes are x=0 and x=-1

the function can't have 0 in the denominator and 0 and -1 will make the denominator 0

the function approaches x=0 and x=-1 but can never touch these lines

change the numerator and denominator to:

(x^2-4x+3)/(2x^2+2x)

the horizontal asymptotes appear as x extends to positive or negative infinity, therefore this fraction approaches what number as x gets very large ?

pick the terms with the greatest exponents in the denominator and the numerator...

x^2/2x^2

when x approaches infinity the other terms will have no bearing on the value of the function

the function will get closer and closer to 1/2 where 1 is the coefficient of the numerator and 2 is the coefficient of the denominator because the "dominant terms" have the same degree of 2 so just divide out the x^2 leaving 1/2

you write the horizontal asymptote as y=1/2

generally speaking...

look at the dominant terms, the terms with the greatest exponents, in the numerator and the denominator

smaller degree/larger degree means y=0 is the horizontal asymptote because the function approaches 0 because the denominator gets larger and larger

same degree/same degree means y=coefficient of numerator/coefficient of denominator

larger degree/smaller degree means no horizontal asymptote because the function approaches infinity

Mehrose F.

09/14/15