Rational Function

Question

given f(x)= t³-2t²-8t/t²-4  find the domain, asymptotes, holes and intercepts.

 

I know that to find the x intercept you set y=0 and solve for x but I am unsure how to isolate x when the quadratic function cannot be used.

From my understanding it is 0=t³-2t²-8t I am just unsure how to solve for x.

 

Any help is appreciated, thank you!

Answer 1

Nice answer by Michael.

We sometimes talk about slant asymptotes when the power of the numerator is one greater than the numerator.

In the case of this problem, since there is a quadratic but no linear term in the denominator, it is simply

t - 2

Answer 2

First, we factor out the numerator and denominator completely.

 

f(x) = [t(t² - 2t - 8)] / [(t - 2)(t + 2)] =

 

f(x) = [t(t - 4)(t + 2)] / [(t - 2)(t + 2)]

 

 

To find the y-intercept, evaluate f(0).

 

To find the hole, find a factor that is in the numerator and denominator.  From this function, you will see that (t + 2) is a factor that is both in the numerator and denominator.  In other words, these two factors cancel each other out.  Therefore, a hole occurs at   t=-2.

 

 

To find the x-intercept, set the numerator equal to zero and solve for t.  Recall that we cancelled out a term in the earlier step.  So now the function becomes

 

f(x) = t(t - 4) / (t - 2)

 

 

Now you can set the numerator equal to zero and solve for t to find the x-intercepts.

 

To find the vertical asymptotes, set the denominator equal to zero and solve for t.

 

There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator.