SOLUTION: Help please Find the slant asymptote,the vertical asymptote,the x and the y intercepts and then sketch the graph of f(x)=(x^3+x^2)/x^2-4

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Question 1090285: Help please
Find the slant asymptote,the vertical asymptote,the x and the y intercepts and then sketch the graph of
f(x)=(x^3+x^2)/x^2-4
Found 2 solutions by KMST, MathLover1:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
IF YOU MEANT f%28x%29=%28x%5E3%2Bx%5E2%29%2F%28x%5E2-4%29 :
Factoring, we get
f%28x%29=%28x%2B1%29x%5E2%2F%28%28x%2B2%29%28x-2%29%29
The function is undefined for x=-2 and x=2 ,
and highlight%28x=-2%29 and highlight%28x=2%29 are the equations of the vertical asymptotes.
The value of the function is zero for x=0 and x=-1 ,
so highlight%28x=-1%29 and highlight%28x=0%29 are the x-intercepts, where y=0 ,
and y=0 is the y-intercept, where x=0 .
In other words, we know the graph goes though (0,0) and (-1,0).
The numerator of the function, %28x%2B1%29x%5E2 , changes sign only at x=-1 .
The denominator of the function, %28x%2B2%29%28x-2%29 , changes sign at x=-2 and x=2 .
All four factors are positive for x%3E2%3E-1%3E-2 , nd so is f%28x%29 .
As you cross x=2 , the fucntion changes sign,
so f%28x%29%3C=0 for -1%3Cx%3C2 , with f%280%29=0 for x=0 .
For -2%3Cx%3C-1 , f%28x%29%3E0 ,
and for x%3C-2 , f%28x%29%3C0 .
As for a slant asymptote, doing the division we find that
%28x%5E3%2Bx%5E2%29%2F%28x%5E2-4%29=x%2B1%2B%284x%2B4%29%28x%5E2-4%29 ,
so highlight%28y=x%2B1%29 is the slant asymptote.
This is what the graph of the function and the slant asymptote looks like:
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C%28x%5E3%2Bx%5E2%29%2F%28x%5E2-4%29%2Cx%2B1%29 .
With the vertical asymptotes added, it would look like this:
.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=%28x%5E3%2Bx%5E2%29%2F%28x%5E2-4%29 -> form f%28x%29+=+p%28x%29+%2F+q%28x%29
you are given a rational function written as the ratio of two polynomials where the denominator q%28x%29 isn't zero
Vertical Asymptotes:
An asymptote is a line that the curve approaches but does not cross. The equations of the vertical asymptotes can be found by finding the roots of q%28x%29. Completely ignore the numerator when looking for vertical asymptotes, only the denominator+matters.
x%5E2-4=0
x%5E2=4
x=sqrt%284%29
x=+2 or x=-2
so, vertical asymptotes: x=+2 and x=-2
Horizontal Asymptotes:
The location of the horizontal asymptote is determined by looking at the degrees of the numerator (n) and denominator (m).
If n%3Cm, the x-axis, y=0 is the horizontal asymptote.
If n=m, then y=an+%2F+bm is the horizontal asymptote. That is, the ratio of the leading coefficients.
If n%3Em, there is no horizontal asymptote.
However, if n=m%2B1, there is an oblique or slant asymptote.
in your case,n=3 and m=2; so, n%3Em which means there is no horizontal asymptote
Oblique Asymptotes:
When the degree of the numerator is exactly one more than the degree of the denominator, the graph of the rational function will have an oblique (slant) asymptote.
in your case n=3 and m=2 and n=m%2B1; so, you have a slant asymptote
To find the equation of the oblique asymptote, perform long division (synthetic if it will work) by dividing the denominator into the numerator.
f%28x%29=%28x%5E3%2Bx%5E2%29%2F%28x%5E2-4%29
------------highlight%28x%2B1%29
+%28x%5E2-4%29/x%5E3%2Bx%5E2
-------------x%5E3-4x
----------------0%2Bx%5E2
-----------------x%5E2-4
---------------------4
so, f%28x%29=%28x%5E3%2Bx%5E2%29%2F%28x%5E2-4%29 is asymptotic to f%28x%29=highlight%28x+%2B+1%29