Problem 1: For each of the rational functions below, follow...

Problem 1: For each of the rational functions below, follow the steps: factor/factor, divide, sign diagrams, graph. Show and explain each of your steps. Write the equations for all asymptotes and/or holes. r(x) = (x^3 - x^2 - x + 1) / (x^2 + x - 2) r(x) = (x^2 - 1) / (x^3 + 4x^2 + 5x + 2) Problem 2: Solve each equation. Express the solution in set notation. Points will be awarded for the details of the algebra and a discussion of the concepts. (1 - x) / x = (x^2 - 1) / x^2 (k + 1) / k = 1 - (k^2 - 3k - 4) / 4k

Transcript

00:01 So, in the first part of the problem 1, it is given that r of x is equal to x cube minus x square minus x plus 1 whole divided by x square plus x minus 2.

00:21 So, we will factorize both the numerator as well as the denominator and we get x square minus 1 multiplied with x minus 1 whole divided by x plus 2 multiplied with x minus 1.

00:37 So, we will further simplify x square minus 1 divided by x plus 2.

00:45 Now, the function r x will have an asymptote, will have an asymptote for x plus 2 is equal to 0 that is x equals to minus 2.

01:03 Therefore, x equals to minus 2 is the answer and x equals to minus 2 is a vertical asymptote, vertical asymptote.

01:23 And the slant asymptote, so for finding the slant asymptote, we can write r x in the form of x minus 2 plus 3 divided by x plus 2.

01:38 So, we see that y equals to x minus 2 is the slant asymptote, asymptote.

01:54 So, this is also our answers.

01:58 Now we will go to the second part of problem 1.

02:03 So, we will put a simple division and we will write r of x in this part is given by x square minus 1 x cube plus 4 x square plus 5 plus 2, we can simplify it in the form of x plus 1 multiplied with x minus 1.

02:29 So, basically we will just factorize both the numerators and the denominators and the denominator we get x plus 1 whole square multiplied with x plus 2, we will further simplify it and we get x minus 1 divided by x plus 1 multiplied with x plus 2.

02:50 So, we see that r x will have an asymptote, asymptote for x plus 1 multiplied with x plus 2 that is the denominator is equal to 0.

03:06 So, this gives us the values x equals to minus 1 and x equals to minus 2 are the vertical asymptotes.

03:24 Why it is vertical asymptote? because we see that the value of x is constant.

03:29 So, the equation of the asymptotes that is x minus x equals to minus 1 is a line parallel to the y axis and x equals to minus 2 is another line parallel to the y axis and this is our answer to the second part of the first problem...

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