# Calculus Summer Assignments

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! Related Introductory Physics Homework Help News on Phys.org
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.

EnumaElish
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.

hotrocks007 said:
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

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.
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

HallsofIvy
Homework Helper
"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 ?