35. For problem 35, you can use the intercepts you’re given to get to the point , because you know that if you solved for each of the values you would end up with the horizontal intercepts given to you in the problem. The problem tells you at what intercepts has what roots of multiplicity to give a degree of 5, which is why is squared, and is cubed. To solve for , (your stretch factor, in this case,), you can plug the point your given, (in this case it’s (2,15)) into your equation: , to solve for . 37. 39. 41. 43. 45. 47. 49. 51. See the diagram below. The area of the rectangle is , and , so Using technology, evaluate the maximum of The -value will be maximum area, and the -value will be half of base length. Dividing the -value by the -value gives us the height of the rectangle. The maximum is at . So, Section 3.4 Solutions 1. b 3. a 5. Vertical asymptote: 4, Horizontal asymptote: , Vertical intercept: , Horizontal intercept: . For problem 5, the vertical asymptote is because that gives a 0 in the denominator, which makes the function undefined. The horizontal asymptote is , because since the degrees are equal in the numerator, and denominator, the ratio of the leading coefficients the answer. The vertical intercept is because , and the horizontal intercept is , because when the function , . 7. Vertical asymptote: , Horizontal asymptote: , Vertical intercept: , Horizontal intercept: none. For this problem, the vertical asymptote is because , a 0 in the denominator, which is undefined. The horizontal asymptote is , because there is only an in the denominator, so as .
35. For problem 35, you can use the intercepts you’re given to get to the point , because you know that if you solved for each of the values you would end up with the horizontal intercepts given to you in the problem. The problem tells you at what intercepts has what roots of multiplicity to give a degree of 5, which is why is squared, and is cubed. To solve for , (your stretch factor, in this case,), you can plug the point your given, (in this case it’s (2,15)) into your equation: , to solve for . 37. 39. 41. 43. 45. 47. 49. 51. See the diagram below. The area of the rectangle is , and , so Using technology, evaluate the maximum of The -value will be maximum area, and the -value will be half of base length. Dividing the -value by the -value gives us the height of the rectangle. The maximum is at . So, Section 3.4 Solutions 1. b 3. a 5. Vertical asymptote: 4, Horizontal asymptote: , Vertical intercept: , Horizontal intercept: . For problem 5, the vertical asymptote is because that gives a 0 in the denominator, which makes the function undefined. The horizontal asymptote is , because since the degrees are equal in the numerator, and denominator, the ratio of the leading coefficients the answer. The vertical intercept is because , and the horizontal intercept is , because when the function , . 7. Vertical asymptote: , Horizontal asymptote: , Vertical intercept: , Horizontal intercept: none. For this problem, the vertical asymptote is because , a 0 in the denominator, which is undefined. The horizontal asymptote is , because there is only an in the denominator, so as .
Chapter%203%20Solutions.docx