Calculus Summer Assignments


by hotrocks007
Tags: assignments, calculus, summer
hotrocks007
hotrocks007 is offline
#1
Aug3-05, 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(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!
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whozum
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#2
Aug3-05, 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.
EnumaElish
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#3
Aug3-05, 11:56 PM
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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
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#4
Aug4-05, 02:06 AM
P: 36

Calculus Summer Assignments


Quote Quote by 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



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
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#5
Aug4-05, 06:12 AM
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Thanks
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"Solve for y:
x^2 + y^2 + 4x - 6y +12=0"

Treat it as a quadratic in y: y2- 6y+ (x[sup]2[sup]+ 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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