
#1
Aug305, 10:52 PM

P: 10

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(y6) 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 (y3)^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
Aug305, 11:15 PM

P: 2,223

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. 



#3
Aug305, 11:56 PM

Sci Advisor
HW Helper
P: 2,483

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. 



#4
Aug405, 02:06 AM

P: 36

Calculus Summer Assignmentsthe 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)(x5)x now just distribute, and you got yourself a cubic function that goes through those zeros 



#5
Aug405, 06:12 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

"Solve for y:
x^2 + y^2 + 4x  6y +12=0" Treat it as a quadratic in y: y^{2} 6y+ (x[sup]2[sup]+ 4x+ 12)= 0 thinking of x^{2}+ 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= x^{2}+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))(x0)(x5)= x(x5)(x+5)= x^{3} 25x ? 


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