course MTH 158 04/17 around 11:00 Your solution, attempt at solution:
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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 #$&* ********************************************* Question: * 5.2.43 / 7th edition 4.3.43. Find the vertical and horizontal asymptotes, if any, of H(x)= (x^4+2x^2+1) / (x^2-x+1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x^4+2x^2+1) / (x^2-x+1) (x^2 + 1)^2 / (x-1)^2 The denominator is zero if x = 1. The numerator is not zero when x = 1 so there is a vertical asymptote at x = 1. no horizontal asymptotes confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The function (x^4+2x^2+1) / (x^2-x+1) factors into (x^2 + 1)^2 / (x-1)^2. The denominator is zero if x = 1. The numerator is not zero when x = 1 so there is a vertical asymptote at x = 1. 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, whose graph is a parabola with vertex at (-1/2, 7/4), opening upward. 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 and that of y = x^2 + x - 1: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: * 5.2.50 / 7th edition 4.3.50. Find the vertical and horizontal asymptotes, if any, of R(x)= (6x^2+x+12) / (3x^2-5x-2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R(x)= (6x^2+x+12) / (3x^2-5x-2) (6•x^2 + x + 12)/((x - 2)•(3•x + 1)) vertical asymptotes at x = 2 and x = -1/3 horizontal asymptote y = 6 x^2 / (3 x^2) = 2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* Overall Understood Material"