SOLUTION: Hi Lovely Tutors, 1.how i can find the Intercepts with asymptotes of: f(x)=f(x)=x^3+2x^2-x-2/x^2+x-6 2.How can i find the zeros of the f(x) and their multiplicity from: f(x)=f

Algebra ->  Rational-functions -> SOLUTION: Hi Lovely Tutors, 1.how i can find the Intercepts with asymptotes of: f(x)=f(x)=x^3+2x^2-x-2/x^2+x-6 2.How can i find the zeros of the f(x) and their multiplicity from: f(x)=f      Log On


   

Question 279129: Hi Lovely Tutors,
1.how i can find the Intercepts with asymptotes of: f(x)=f(x)=x^3+2x^2-x-2/x^2+x-6
2.How can i find the zeros of the f(x) and their multiplicity from: f(x)=f(x)=x^3+2x^2-x-2/x^2+x-6
I also want to thank all tutors, thanks for all the help so far and am sorry to ask so much questions, but i love math and want to know all i can, thanks to all of you i am understanding exactly how to solve what i didn't know:)
If any of the Great tutors can help me out than that be the greatest help i can get right now, and i will surely replay with a comment to the tutor as always:)
Thank you:)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=%28x%5E3%2B2x%5E2-x-2%29%2F%28x%5E2%2Bx-6%29
A rational function is a fraction where both the numerator and denominator are polynomials. When analyzing and/or graphing rational functions the zeros of the two polynomials tell you most of what you need to know. And since zeros of polynomials are found by factoring them, factoring is a huge part of analyzing and/or graphing rational functions.

So we start by factoring the numerator and denominator. The denominator is a trinomial that factors very easily:
f%28x%29=%28x%5E3%2B2x%5E2-x-2%29%2F%28%28x%2B3%29%28x-2%29%29
The numerator will factor by grouping:
f%28x%29=%28x%5E2%28x%2B2%29+%2B+%28-1%29%28x%2B2%29%29%2F%28%28x%2B3%29%28x-2%29%29
f%28x%29=%28%28x%2B2%29%28x%5E2+%2B+%28-1%29%29%29%2F%28%28x%2B3%29%28x-2%29%29
We can continue to factor the numerator using the difference of squares pattern:
f%28x%29=%28%28x%2B2%29%28x%2B1%29%28x-1%29%29%2F%28%28x%2B3%29%28x-2%29%29
The function is now fully factored. And in factored form we can see (or figure out) the values of x that make a factor zero. And if a factor is zero, the product is zero.

The values of x that make the numerator zero are -2, -1 and 1. The values of x that make the denominator zero are -3 and 2.

If you can't "see" these numbers in the factors, then set each factor equal to zero and solve the equation. For example, for the factor x+2:
x+2 = 0
Subtract 2 from each side:
x = -2

Now we're ready to figure out the details of your function:
  1. "Holes" (I think the technical term is "removable discontinuity") occur for x values that make both the numerator and denominator zero. (SPECIAL CASE: If every factor of the denominator represents a "hole" then your function is a special case. An explanation of how to handle this special case is at the end.) There are no "holes" in your function.
  2. Vertical asymptotes occur for x values that make just the denominator zero. So your function will have vertical asymptotes at x = -3 and x = 2.
  3. X-intercepts occur for x values that make just the numerator zero. So your function will have x-intercepts at -2, -1 and 1.

For the rest of what we need, we'll use the original, non-factored form of the function, f%28x%29=%28x%5E3%2B2x%5E2-x-2%29%2F%28x%5E2%2Bx-6%29:
  • The y-intercept occurs when x = 0. This can be found easily because if x is zero any term with x in it will be zero, too. Looking at your function we should be able to see that if x is zero, the numerator becomes -2 and the denominator becomes -6 making the function have a value of -2/-6 or 1/3 when x is zero. So the y-intercept is 1/3.
  • Horizontal, oblique and non-linear asymptotes. These are determined in large part by the degrees (highest exponents) of the numerator and denominator:
    • If the degree of the numerator is less than the degree of the denominator, then the line y = 0 will be a horizontal asymptote.
    • If the degrees are equal, then there will be a horizontal asymptote at the value of the ratio of leading coefficients (the coefficients of the terms with the highest exponent). For example, the function h%28x%29+=+%283x%5E6+-15x+%2B4%29%2F%282x%5E6+-200x%5E5+%2B45x+-3%29 has leading coefficients of 3 and 2. Its horizontal asymptote will be y = 3/2.
    • If the degree of the numerator is greater than the degree of the denominator, then there will not be a horizontal asymptote. But there will be either an oblique or a non-linear asymptote. (These are not always taught so you may not be responsible for finding them.) The degree of your numerator is 3 and the degree of your denominator is 2. So you will have a either an oblique or a non-linear asymptote.
    The logic behind all of this is based on what happens to fractions when x becomes very large positive or negative numbers.

Finding an oblique or a non-linear asymptote:
  1. Divide the fraction.
  2. Discard the remainder. (If there is no remainder, then you missed the SPECIAL CASE mentioned above under "Holes" and explained below.)
  3. The remaining equation is an asymptote. If it is the equation of a line, then it is called an oblique asymptote. Otherwise it is a non-linear asymptote.

Since your function has this type of asymptote we will work through these steps on your function:
1. Divide
          x    + 1
          ________________________________
x^2+x-6  /x^3  +2x^2  -x   -2
          x^3  + x^2  -6x
          ----------------
                 x^2  +5x  -2
                 x^2  + x  -6
                 ------------
                       4x  +4

So

2. Discard the remainder.
(Since this changes the equation I will no longer call it f(x)):
y = x + 1
3. So y = x + 1 is an asymptote. This is the equation of a line so it is called an oblique asymptote.

Here's a look at the graph of your function. (The obliques asymptote is in green. The vertical asymptotes, x = -3 and x = 2, are not drawn because I do not know how to get Algebra.com's software to draw them. I hope they are obvious.):


**SPECIAL CASE. If every factor of the denominator represents a "hole":
  1. Cancel the common factors in the numerator and denominator.
  2. You now have a function which is not a rational function. The graph of the original function will be the same as the graph of this new, non-rational function except there will be be "holes" in it for the x values that make the original denominator zero.

For example, if g%28x%29+=+%28%28x%2B4%29%28x-3%29%28x%2B7%29%29%2F%28%28x-3%29%28x%2B4%29%29 then, after canceling, we have just x+7. y = x+7 is a line. The graph of g(x) will be this same line except that there will be "holes" at x = 3 and x = -4.