Calculus Summer Assignments

  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. 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. EnumaElish

    EnumaElish 2,332
    Science Advisor
    Homework Helper

    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. 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. HallsofIvy

    HallsofIvy 41,260
    Staff Emeritus
    Science Advisor

    "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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