SOLUTION: Please help me with the equation below: Identify the intercepts, asymptotes, and "holes" on this rational equation: f(x) = x^3+4x^2-5x/x^2-1

Algebra ->  Rational-functions -> SOLUTION: Please help me with the equation below: Identify the intercepts, asymptotes, and "holes" on this rational equation: f(x) = x^3+4x^2-5x/x^2-1      Log On


   



Question 983779: Please help me with the equation below:
Identify the intercepts, asymptotes, and "holes" on this rational equation:
f(x) = x^3+4x^2-5x/x^2-1

Found 2 solutions by solver91311, KMST:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


1. Find the -intercept. Substitute zero for . If the function is defined at , then the value of the function at is the -coordinate of the -intercept. Remember that a -intercept has the form

2. Find the -intercept(s). Set the numerator of the rational function equal to zero and solve for all real number roots. Each real number root of the numerator will be an -coordinate of an -intercept. Remember that an -intercept has the form

3. Find any vertical asymptotes or holes. Factor both the numerator and denominator polynomials completely. If there is a factor that is in both the numerator and denominator, the value of when that factor is set equal to zero is the location of a "hole". If there is a factor in the denominator that is not in the numerator, the value of when that factor is set equal to zero is the location of a vertical asymptote.

4. Find any horizontal or oblique asymptotes.
If the degree of the numerator is smaller than the degree of the denominator, there is a horizontal asymptote at

If the degree of the numerator is equal to the degree of the denominator, then there is a horizontal asymptote at where is the lead coefficient of the numerator polynomial and is the coefficient of the denominator polynomial.

If the degree of the numerator is one greater than the degree of the denominator then there is an oblique asymptote. Using Polynomial Long Division, find the quotient of the numerator divided by the denominator, excluding any remainder. This quotient is the oblique asymptote function descriptor.

If you are unsure of the process for Polynomial Long Division, see: Purple Math Polynomial Long Division

John

My calculator said it, I believe it, that settles it

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I assume you mean f%28x%29=%28x%5E3%2B4x%5E2-5x%29%2F%28x%5E2-1%29 .
The y-intercept is the point where the graph crosses the y-axis,
and for all points on the y-axis x=0 ,
so the y-intercept is f%280%29=0%2F%28-1%29=0 .
The function crosses the y-axis at the point with system%28x=0%2Cy=0%29 ,
The highlight%28y-intercept%29 is the point (0,0), the origin.
Factoring numerator and denominator, we get
.
For x%3C%3E1 , the function is not defined, because the denominator is zero.
However for values of x around x=1 f%28x%29 has a value.
There is just a highlight%28hole%29 at x=1 .
Otherwise, for x%3C%3E1 , the expression simplifies to
f%28x%29=x%28x%2B5%29%2F%28x%2B1%29 ,
When a factor in the x%28x%2B5%29 numerator is zero, f%28x%29=0 ,
and the graph crosses the x-axis, with y=0
The points where that happens are the highlight%28x-intercepts%29 :
highlight%28x=0%29 , and
x%2B5=0--->highlight%28x=-5%29 .
So the graph crosses the x-axis at (0,0) (we knew that point) and at (-5,0).
Also, for x%3C%3E1 , we see that f%28x%29=x%28x%2B5%29%2F%28x%2B1%29 is undefined for x=-1 , because the denominator is zero.
For values of x very close to x=-1 ,
the numerator, the quadratic polynomial , and
as x approaches -1 , P%28x%29 approaches -4 .
On the other hand, as x approaches -1 ,
the denominator, x%2B1 , approaches zero.
As a consequence, the absolute value of the quotient f%28x%29=x%28x%2B5%29%2F%28x%2B1%29 grows without bounds.
That makes highlight%28x=-1%29 a highlight%28vertical%29highlight%28asymptote%29 .
Also, for .
and f%28x%29-%28x%2B4%29=-4%2F%28x%2B1%29 approaches zero as abs%28x%29 increases.
That makes highlight%28y=4x%2B4%29 an highlight%28asymptote%29 .
With all that information, we can sketch the graph of f%28x%29 ,
which looks like this

NOTE:
f%28x%29=%28x%5E3%2B4x%5E2-5x%29%2F%28x%5E2-1%29 should be written as f(x)=(x^3+4x^2-5x)/(x^2-1),
because according with the algebra conventions known as "order of operations" rules,
multiplication and divisions are done before addition and subtraction, so
x^3+4x^2-5x/x^2-1 = x%5E3%2B4x%5E2-5x%2Fx%5E2-1 .
The long horizontal line separating x%5E3%2B4x%5E2-5x from x%5E2-1 is a very efficient grouping symbol:
it saves you from writing two pairs of parentheses.
However, when you have to key in symbols one at a time,
you have to key in all those parentheses that are implied.
Otherwise, you can get in trouble.
%280.735-0.004%29%2F%280.720-0.008%29=0.731%2F0.712=1.027 (rounded to 3 decimal places).
Most calculators know about order of operations,
and if you key in
0.735 - 0.004 / 0.720 - 0.008 =0.735-0.004%2F0.720-0.008 ,
you will get 0.735-0.004%2F0.720-0.008=0.735-%220.00555...%22-0.008=0.735-%220.01355...%22=0.721 (rounded to 3 decimal places).