Question 199342: Give the equations of any vertical,horizontal, or oblique asymptotes.
1.f(x)= 1/x-2
2.f(x)= x+2/x-2
3.f(x)= x^2+x-2/x-2
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! Remember,,, form y= a(n)x^n +..... / b(m) x^m +......
.
for VERTICAL ASYMPTOTES,,,,,,find zero"s of denominator
.
in all 3 cases above,,, (x-2) =0,,,or x=2 is a vertical asymptote
.
For HORIZONTAL ASYMPTOTES,,,,
.
if n less than m,,,,,,y=0
.
if n=m,,,,,y = a(n)/b(m)
.
n>m,,,,,,, no horizontal asymptote
.
In the 1st case,,, n=0,m=1,,,or n is less than m,,,,,y=0 is hor asymptote
.
in the 2nd case,,,n=1,m=1,,or n=m,,,,y = a(n)/b(m) = 1/1 =1,,,, y=1 is hor asymptote
.
in the 3rd case, n=2,m=1 ,,or n>m,,,,,therefore no hor asymptote
.
for OBLIQUE ASYMPTOTES (SLANT)
.
if n=(m+1),,,divide numerator by denominator ,,,,and set answer (without remainder) =y
.
in the 3rd case above ,,,n=2,m=1,,or n=(m+1),,,,therefore there is a SLANT asymptote
.
( x^2 +x -2) / (x-2) = x+3 +4/(x-2)
.
setting y = x+3,,,,,,for slant asymptote
.
.We might also look for x intercepts,,,,which are zero's of numerator.
.
case 2 ,,,has one at x=-2
.
case 3,,,has one at x=-2,,and another at x=+1
.
|
|
|
