Answer :
The horizontal asymptotes will occur when x approaches ±oo
The easy way to see this is to divide all terms by the highest order variable...
(x^2/x^3 +x/x^3-6/x^3)/(x^3/x^3-1/x^3) and all of that mess boils down to
0/1 or just 0 as x approaches ±oo
So the horizontal asymptote is the horizontal line y=0
The vertical asymptote will occur when there is division by zero, which is undefined because it is not a real value.
x^3-1=0
(x-1)(x^2+x+1)=0
So this only occurs as x approaches -1, so the vertical asymptote is the vertical line x=-1.
The easy way to see this is to divide all terms by the highest order variable...
(x^2/x^3 +x/x^3-6/x^3)/(x^3/x^3-1/x^3) and all of that mess boils down to
0/1 or just 0 as x approaches ±oo
So the horizontal asymptote is the horizontal line y=0
The vertical asymptote will occur when there is division by zero, which is undefined because it is not a real value.
x^3-1=0
(x-1)(x^2+x+1)=0
So this only occurs as x approaches -1, so the vertical asymptote is the vertical line x=-1.
Answer:
Hence, x=1 is the vertical asymptote
Hence, y=0 is the horizontal asymptote
Step-by-step explanation:
We have been given an expression
[tex]\frac{x^2+x-6}{x^3-1}[/tex]
For vertical asymptote we equate the denominator to zero that means
[tex]x^3-1=0[/tex]
[tex]\Rightarrow x^3=1[/tex]
[tex]\Rightarrow x=1[/tex]
Hence, x=1 is the vertical asymptote.
Now, for horizontal asymptote we compare the degree of numerator and denominator
Here degree of numerator is 2 and degree of denominator is 3
When degree of denominator is greater than degree of numerator y=0 is the horizontal asymptote
Hence, y=0 is the horizontal asymptote.