Question 279129
{{{f(x)=(x^3+2x^2-x-2)/(x^2+x-6)}}}
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.<br>
So we start by factoring the numerator and denominator. The denominator is a trinomial that factors very easily:
{{{f(x)=(x^3+2x^2-x-2)/((x+3)(x-2))}}}
 The numerator will factor by grouping:
{{{f(x)=(x^2(x+2) + (-1)(x+2))/((x+3)(x-2))}}}
{{{f(x)=((x+2)(x^2 + (-1)))/((x+3)(x-2))}}}
We can continue to factor the numerator using the difference of squares pattern:
{{{f(x)=((x+2)(x+1)(x-1))/((x+3)(x-2))}}}
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.<br>
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.<br>
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<br>
Now we're ready to figure out the details of your function:<ol><li>"Holes" (I think the technical term is "removable discontinuity") occur for x values that make <b>both</b> the numerator and denominator zero. (SPECIAL CASE: If <b>every</b> 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.</li><li>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.</li><li>X-intercepts occur for x values that make just the numerator zero. So your function will have x-intercepts at -2, -1 and 1.</li></ol>
For the rest of what we need, we'll use the original, non-factored form of the function, {{{f(x)=(x^3+2x^2-x-2)/(x^2+x-6)}}}:<ul><li>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.</li><li>Horizontal, oblique and non-linear asymptotes. These are determined in large part by the degrees (highest exponents) of the numerator and denominator:<ul><li>If the degree of the numerator is less than the degree of the denominator, then the line y = 0 will be a horizontal asymptote.</li><li>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(x) = (3x^6 -15x +4)/(2x^6 -200x^5 +45x -3)}}} has leading coefficients of 3 and 2. Its horizontal asymptote will be y = 3/2.</li><li>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.</li></ul>The logic behind all of this is based on what happens to fractions when x becomes very large positive or negative numbers.</li></ul>
Finding an oblique or a non-linear asymptote:<ol><li>Divide the fraction.</li><li>Discard the remainder. (If there is no remainder, then you missed the SPECIAL CASE mentioned above under "Holes" and explained below.)</li><li>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.</li></ol>
Since your function has this type of asymptote we will work through these steps on your function:
1. Divide
<pre>
          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
</pre>
So {{{f(x)=(x^3+2x^2-x-2)/(x^2+x-6) = x + 1 + (4x + 4)/(x^2+x-6)}}}<br>
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.<br>
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.):
{{{graph(400, 400, -10, 10, -10, 10, (x^3+2x^2-x-2)/(x^2+x-6), (sqrt(sin(5x))/sqrt(sin(5x)))(x+1))}}}<br>
**SPECIAL CASE. If every factor of the denominator represents a "hole":<ol><li>Cancel the common factors in the numerator and denominator.</li><li>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 <i>except there will be be "holes" in it for the x values that make the original denominator zero.</i></li></ol>
For example, if {{{g(x) = ((x+4)(x-3)(x+7))/((x-3)(x+4))}}} then, after canceling, we have just x+7. y = x+7 is a line. The graph of g(x) will be this same line <i>except that there will be "holes" at x = 3 and x = -4.</i>