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Homework Help: Calculus Summer Assignments

  1. Aug 3, 2005 #1
    I have a summer calc assignment due tomorrow, and I've been stuck on these two problems. Help is appreciated!!!

    Solve for y:
    x^2 + y^2 + 4x - 6y +12=0
    I tried this:
    x^2 + 4x + 12 = -y(y-6)
    But I don't think that was correct. THe left side wont' factor, and I dont htink that is the right way to approach it.

    I also tried to complete the square:
    y^2 - 6y = -x^2 - 4x - 12
    y^2 - 6y + 9 = -x^2 -4x -12 +9
    (y-3)^2 = -x^2 - 4x - 3
    But I was not sure how to take the square root of the right side.

    Also, this problem was to create an equation whose graph has intercepts at (-5,0), (0,0), and (5,0). I tried a sine graph, but it all seems like guess and check, and I cannot get the intercepts at 5 and -5 exactly.

    Thanks so much!
  2. jcsd
  3. Aug 3, 2005 #2
    The square root of the right side will most likely not simplify nicely. Completeing the square is the best way in my opinion.

    For the second one, I think you can modify a sin wave to have those intercepts, but I would try piecewise for the easiest way out.
  4. Aug 3, 2005 #3


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    Let C = x^2 + 4x + 12,
    B = -6
    A = 1
    Then Ay^2 + By + C = 0. Now use the quadratic root formula to solve for roots.

    How can this polynomial have a (0,0) intercept? At (0,0) left side = 12.
  5. Aug 4, 2005 #4
    first one... just complete the square, at least that's what i'd do

    the second question:
    really simple trick
    you want all those points to basically be zero's correct?

    thus, -5, 0, and 5 would be zero's of this equation.... i'm thinking:


    now just distribute, and you got yourself a cubic function that goes through those zeros
  6. Aug 4, 2005 #5


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    "Solve for y:
    x^2 + y^2 + 4x - 6y +12=0"

    Treat it as a quadratic in y: y2- 6y+ (x2+ 4x+ 12)= 0 thinking of x2+ 4x+ 12 as the constant term. Either complete the square (you will, of course, be left with a square root) or use the quadratic formula with a= 1, b= -6, and c= x2+4x+12.

    "Also, this problem was to create an equation whose graph has intercepts at (-5,0), (0,0), and (5,0)."

    The equation has 0 values at -5, 0, 5. How about y= (x-(-5))(x-0)(x-5)= x(x-5)(x+5)= x3- 25x ?
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