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!
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.
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.
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: (x+5)(x-5)x now just distribute, and you got yourself a cubic function that goes through those zeros
"Solve for y: x^2 + y^2 + 4x - 6y +12=0" Treat it as a quadratic in y: y^{2}- 6y+ (x^{2+ 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 ?}