T3 answers

Algebra

Fall 2001

MATH 121-11

Test #3

Instructions: Answer all problems correctly. NO CALCULATORS! Each numbered problem is worth 10 points unless otherwise specified. Each st*rred problem is extra credit, and each * is worth 4 points.

1. Use synthetic division to write the rational function
   

   f(x) =  x (x - 1) x + 1

   in the form
   

   q(x) +  r(x) d(x) ,

   where q(x) is a polynomial and where the degree of r is less than the degree of d. Note: Be particularly careful on this problem and the next, as you can make use of your result on a subsequent problem.
   First, you need to rewrite the function as
   

   f(x) =  x2 - x x + 1 ,

   and also keep in mind that
   

   x2 - x = x2 - x + 0,

   as you need the 0 as a place-holder. Here's the synthetic division tableau:
    
   
    
   The result is that
   

   f(x) = x - 2 +  2 x+1 .

2. Use polynomial long division to write the rational function
   

   f(x) =  x2 (x + 2) (x + 1)2

   in the form
   

   q(x) +  r(x) d(x) ,

   where q(x) is a polynomial and where the degree of r is less than the degree of d.
   First, you need to rewrite the function as
   

   f(x) =  x3 + 2x2 x2 + 2x + 1 .

   Then long division gives us
    
   
    
   Thus,
   

   f(x) = x +  -x x2 + 2x + 1 = x -  x (x + 1)2 .

3. Do one (and only one) of the following.

   1. (15 points) Neatly sketch the graph of the function in problem #2.
      Looking at the formula obtained in problem #2,
      

      f(x) =  x2 (x + 2) (x + 1)2 = x -  x (x + 1)2 .

      we see that f(x) has an oblique asymptote of y = x. (This is the quotient, q(x), in that problem. We also see that there must be a vertical asymptote of x = -1 and two zeros, one at x = 0 the other at x = -2, these being most easily seen from the factored form of the function. (The zero at x = 0 has degree 2, so the local behavior will be parabolic and the graph won't change sign there.)
      
   2. (10 points) Neatly sketch the graph of the function in problem #1.

      Be sure to clearly make your own x- and y-axes. Whichever function you choose, be sure to list all asymptotes and zeros (if any) of the function.

      The function
      

      f(x) =  x (x - 1) x + 1 = x - 2 +  2 x+1

      has an oblique asymptote of y = x - 2, zeros at x=0 and x=1, and a VA at x = -1.
      
   3. (7 points) Neatly sketch the graph of the function
      

      f(x) =  x-1 (x-2)(x+1) .

      The degree of the numerator is less than that of the denominator, so we'll have a horizontal asymptote at y = 0. We have a zero at x = 1 and VA's at x = -1 and x = 2.
      

4. For the function f(x) you selected in problem #3, solve the inequality f(x) £ 0.
   Just looking at the graphs gives us everything; just select the intervals where the function lies on or below the x-axis.

   1. The solution set is {x £ -2 or x = 0}. This can be written as [-¥, -2] È{0}.
   2. {x < -1 or 0 £ x £ 1}   =   (-¥, -1) È[0,1].
   3. {x < -1 or 1 £ x < 2}   =   (-¥, -1) È[1,2).
5. List all of the possible rational zeros of the polynomial
   

   p(x) = 7x3 + a x2 - 6.

   (If you're wondering, ``What's a?'' then you shouldn't be.)
   The a has nothing to do with it, as we need only consider the first and last coefficients. (Actually, I should have said that a is a real number.) The possible rational seros are
   

   ±  {1,2,3,6} {1,7}   =   ±{ 1, 2, 3, 6, 1/7, 2/7, 3/7, 6/7 }

6. Construct a polynomial of degree three with zeros at x=-1, x=-2 and x=3 and having a y-intercept of -12.
   Having degree three and those zeros means the polynomial has x+1, x+2 and x-3 as its only factors. So it must be a multiple of the product of those. Thus it must be
   

   p(x) : = A(x+1)(x+2)(x-3)

   for some constant A. having a y-intercept of -12 is equivalent to having
   

   p(0) = -12.

   Therefore,
   

   p(0) = A(0+1)(0+2)(0-3) = -12,

   or
   

   -6A = -12,

   hence A = 2 and
   

   p(x) = 2(x+1)(x+2)(x-3).

   Some of you thought it was necessary to expand the polynomial. In that case,
   

   p(x) = 2(x3 - 7x - 6) = 2x3 - 14x - 12.

7. By some miracle, you know that 1-2i is a zero of the polynomial
   

   P(x)=2x4-4x3+6x2+8 x-20.

   Find all the other zeros of P(x).
   Since 1-2i is a zero and since the polynomial has only real coefficients, we know by the Conjugate Zeros Theorem that the conjugate 1+2i is also a zero. We can use synthetic division to divide out the factors corresponding to these zeros:
   
   The zeros of the quotient, 2x² - 4, can easily be found to be x = ±Ö2. So the zeros of the polynomial are
   

   {1 + 2i, 1 - 2i, Ö2, -Ö2 }.

   
   Note: There is another trick for reducing the polynomial using long division instead of synthetic division. Since we know the two complex zeros 1 ±2i, we know that the following polynomial is a factor:
   

   (x - (1+2i)) (x - (1-2i)) = ((x-1) - 2i) ((x-1) + 2i) = (x-1)2 + 22 = x2 -2x + 5.

   So divide
   

   2x4-4x3+6x2+8 x-20 x2 -2x + 5

   using long division, and you'll get 2x² - 4.
8. Factor the following polynomial completely.
   

   x4-5 x3+8 x2-5 x+1

   The only possible rational zeros are ±1, so checking is easy.
    
    
    
   Therefore,
   

   x4-5 x3+8 x2-5 x+1 = (x - 1)2 (x2 - 3x + 1).

   If you stopped there, I gave 7 or 8 points, depending on my mood at the time. But this can now be reduced further using the quatratic formula, which gives
   

   3 ±Ö5 2

   as the zeros of x² - 3x + 1. So the complete factorization is
   

   x4-5 x3+8 x2-5 x+1 = (x - 1)2 æ è x -  3 + Ö5 2 ö ø æ è x -  3 - Ö5 2 ö ø .

9. Sketch the ellipse given by
   

   9x2+4y2-36x+24y=-36.

   (Supply your own x and y-axes.)
   First we need to complete the squares. (Many of you still don't do this correctly.)
   

   9x2 + 4y2 - 36x + 24y = -36 9(x2 - 4x) + 4(y2 + 6y) = -36 9(x2 - 4x + 4 - 4) + 4(y2 + 6y + 9 - 9) = -36 9(x2 - 4x + 4) - 36 + 4(y2 + 6y + 9) - 36 = -36 9(x - 2)2 + 4(y + 3)2 = -36 + 36 + 36 9(x - 2)2 + 4(y + 3)2 = 36  (x - 2)2 4 +  (y + 3)2 9 = 1

   So the center of the ellipse is at (h, k) = (2, -3) and the horizontal and vertical semi-axes are a = 2 and b=3, resp.
    
    
   
10. For the ellipse below:

     
    

     
    

    1. Give the equation in standard form;
       By inspection, the center is (h, k) = (-2, 1) and the horizontal and vertical semi-axes are a = 2 and b=4, resp. So the equation is
       

       (x + 2)2 4 +  (y - 1)2 16 = 1

    2. Give the coordinates of the foci. (pronounced fo-sigh, incidentally)
       The distance c from the ellipse's center to a focus is given by
       

       c = Ö   a2 - b2        or      c = Ö   b2 - a2   ,

       whichever gives a real result (the first, if the ellipse is elongated horizontally, the second if vertically). In this case,
       

       c = Ö   42 - 22   = Ö   12   = 2 Ö3.

       So the foci are at
       

       (h, k ±c) = (-2, 1 ±2 Ö3).

11. An ellipse has a focus at (-1,1), center at (0,1) and passes through the point (2,1). Write the equation in standard form for the ellipse.
    Drawing a crude picture helps.  
     
    You can then see that a=2 and c=1, and so, since c² = a² - b², we have
    

    b2 = a2 - c2 = 22 - 12 = 3.

    Therefore,
    

    x2 4 +  (y - 1)2 3 = 1.

12. Find the focus and directrix of the parabola given by y = 2x²-4x.
    I'll rewrite the equation in the standard "vertex" form:
    

    y = 2(x2 - 2x) = 2(x2 - 2x + 1 - 1) = 2(x2 - 2x + 1) - 2 = 2(x - 1)2 - 2.

    So the vertex is at
    

    (h, k) = (1, -2).

    We have that p = 1/4a, where p is the distance from the focus to the vertex.Here a = 2, so p = 1/8. Therefore the focus is 1/8 unit from the vertex (above it, since a > 0) at
    

    (h,k+p) = (1, -2 + 1/8) = (1, -15/8).

    The directrix is p units velow the vertex:
    

    y = k-p = -2 - 1/8 = -17/8.

13. A parabola has its focus at the point (-3,2) and its directrix is the line y=6. Write an equation for the parabola.
    The focus is below the directrix, so the parabola opens downward. The vertex is between the focus and directrix, so lies at
    

    (h, k) = (-3, (2+6)/2) = (-3, 4).

    The distance from focus to vertex is (6-2)/2 = 2 and this is the absolute value of p. So p = -2, hence a = 1/4p = -1/8. The parabola's equation is
    

    y = -  1 8 (x + 3)2 + 4.