Calculus Summer Assignments

[hotrocks007]

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 don't 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!

 

[whozum]

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]

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.

 

[Hessam]

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]

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

Treat it as a quadratic in y: y²- 6y+ (x<sup>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 ?</sup>