If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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* * *  5.2.22 / 5.2.20 / 7th edition 4.3.20. Find the domain of G(x)= (x-3) / (x^4+1).

Your solution:

Confidence Assessment:

Given Solution:

The denominator is x^4 + 1, which could be zero only if x^4 = -1. However x^4 being an even power of x, it cannot be negative. So the denominator cannot be zero and there are no vertical asymptotes.

Self-critique (if necessary):

Self-critique Rating:

Question: *  Extra Problem / 5.2.43 / 7th edition 4.3.43. Find the vertical and horizontal asymptotes, if any, of H(x)= (x^4+2x^2+1) / (x^2-x+1).

Your solution:

Confidence Assessment:

Given Solution:

The function (x^4+2x^2+1) / (x^2-x+1) factors into

(x^2 + 1)^2 / (x^2-x+1).

The denominator cannot be zero, since the quadratic-formula solution to the equation x^2 - x + 1 = 0 contains the expression sqrt( (-1)^2 - 4 * 1 * 1) = sqrt(-3) (i.e., the discriminant is negative), and the square root of a negative number isn't a real number.

So there are no vertical asymptotes.

The degree of the numerator is greater than that of the denominator so the function has no horizontal asymptotes.

STUDENT QUESTION

 So, if the degree of the numerator is greater than that of the denominator, is this saying it will always be no horizontal
asymptotes? Where does the long division come in?

INSTRUCTOR RESPONSE

Long division shows the function to which the graph is asymptotic.

 

The correct quotient is x^2 + x + 2 + (x - 1) / (x^2 - x + 1).

• As x gets large, the fraction (x - 1) / (x^2 - x + 1) approaches 0.

• As a result, the graph of the function approaches that of x^2 + x + 2.  This graph is a parabola with vertex at (-1/2, 7/4), opening upward.

• So the graph is asymptotic to the parabola, meaning that for large x, the graph of the given function approaches that of the parabola.

You don't actually need to know how to deal with this particular situation, but you might need to know about 'slant
asymptotes'. The principle is the same:

Slant asymptotes occur when the degree of the numerator exceeds that of the denominator by 1. For example if the question had been about the function

• f(x) = (x^3+2x^2+1) / (x^2-x+1)

the long division would have given you

• x + 3 + (2x - 2) / (x^2 - x + 1).

Again for large x the fraction would approach zero, this time leaving you with the linear function y = x + 3 (the line with slope 1 and y intercept 3).

 

Whatever else the graph does, as you move to the right or to the left the graph eventually approaches this line.

 

The figure below depicts this function and the graph of y = x + 3:

 

The next figure depicts the graph of the original function y = (x^4+2x^2+1) / (x^2-x+1) and that of y = x^2 + x - 1.  It's not hard to imagine that if you go far enough, the two graphs get closer and closer together.

 

 

Self-critique (if necessary):

Self-critique Rating:

Question: *  Extra Problem:  5.2.50 / 7th edition 4.3.50. Find the vertical and horizontal asymptotes, if any, of R(x)= (6x^2+x+12) / (3x^2-5x-2).

Your solution:

Confidence Assessment:

Given Solution:

* * The expression R(x)= (6x^2+x+12) / (3x^2-5x-2) factors as

(6•x^2 + x + 12)/((x - 2)•(3•x + 1)).

The denominator is zero when x = 2 and when x = -1/3. The numerator is not zero at either of these x values so there are vertical asymptotes at x = 2 and x = -1/3.

The degree of the numerator is the same as that of the denominator so the leading terms yield horizontal asymptote

y = 6 x^2 / (3 x^2) = 2.

Self-critique (if necessary):

Self-critique Rating: