Pages: 1

 

2008-05-17 17:31:29
BigOrangeArmy
Don't tase me, bro!
+12|6467|Dallas
I've got a big project in Algebra (due on Monday), where I have to form an image on my calculator with:

*2 lines that are neither horizontal nor vertical
*1 absolute value
*2 circles
*2 ellipses
*2 parabolas, 1 horizontal and 1 vertical
*1 hyperbola
*At least 2 more conic equations

I have a couple questions (im a bit of'a noob with these ), so...

My calculator, when imputting the equations, cannot create "Y" values. For example, if I wanted a horizontal parabola, it would be impossible because the equation for a horizontal parabola is this:

      ---X= Y^2 + 6y -12.

Another equation, for the circles, goes like this:

      ---X^2 + Y^2 + 5x + 6y + 36.

Is there any way by hand I can change these equations to the "Y=" format? The calculator, by the way, is a TI-83 Plus. Any help is great, cause im in a bit of a pickle here .

Last edited by BigOrangeArmy (2008-05-17 17:32:20)

2008-05-17 18:06:57
argo4
Stand and Deliver
+86|6404|United States
have you tried just solving for y?  for the circle graphs you're gonna end up with 2 y= graphs cause you are gonna be taking square roots.
2008-05-17 18:15:43
Jestar
Shifty's Home Number: 02 9662 8432
+373|7212

solve(X= Y^2 + 6y -12.,y)
2008-05-17 18:24:56
l41e
Member
+677|7119

First of all, the assignment is impossible, at least for you. The TI-83+ has 10 equation slots. If I counted correctly, including problems that need more than one slot, you need 15 equation slots. I'll help you as best I can, anyway, though...

Lines:
Y = X
Y = -X

Absolute value:
Y = abs(X (absolute value is in the "math" key menu somewhere)

Circles and ellipses I forget how to do, but solve the equations for Y. Then, wherever there's a plus/minus sign (remember, put those in whenever you take a square root), make one equation for the plus, and one for the minus.

Parabolas:
Vertical: Y = X2
Horizontal requires two equations: Y = square root of X, and Y = negative square root of X.

Hyperbola:
Y = 5/X

Other conic equations:
I suppose you can probably do these yourself by adapting the above equations. Just mess with the numbers (not the variables) and put pluses and minuses to move stuff around. For example, Y = 1/(X+2) will make a "thinner" hyperbola and move it...somewhere, I don't know. I forget.

Good luck.

Last edited by k30dxedle (2008-05-17 18:26:44)

2008-05-17 18:32:35
BigOrangeArmy
Don't tase me, bro!
+12|6467|Dallas

argo4 wrote:

have you tried just solving for y?  for the circle graphs you're gonna end up with 2 y= graphs cause you are gonna be taking square roots.
I've tried. With this equation: X= Y^2 + 6y -12,

I went to: x+12=Y^2+6y
    *Complete the Square: (6/2)^2=9, so---  X + 12 + 9= Y^2 + 6y + 9

And.... (x+21)y+3)^2. Thanks . But how do I solve it from there to the "Y=" format? Do I use the Square Root of both sides?

Im having problems with the circle though. With the Circle Equation, I've got from:

     ---X^2 + Y^2 + 5x + 6y + 36

to:

     ---(X^2+5x) + (Y^2+6y) = -36

Do I complete the square again to get it into standard form? Or do I need to distribute it? Either way, I need to put it into a "Y=" format so that I can plug it into the calculator...
2008-05-17 18:38:01
argo4
Stand and Deliver
+86|6404|United States
for the X^2+Y^2+5X+6Y+ 36=0 you end up with (x+2.5)^2 + (y+3)^2 =-20.75 in the normal form (x-h)^2 +(y-k)^2 =r^2  This graph is not possible cause you can't square a positive radius and get a negative number like -20.75 .

Try another graph like x^2 +y^2 + 4x -6y-3 =0



Edit:
x²+y²+4x-6y-3=0
x²+4x+ y²-6y=3
(x²+4x+4) + (y²-6y+9) =3+9+4  complete square
(x+2)² +(y-3)²= 16   standard form
(y-3)² = 16 -(x+2)²
(y-3)² =16 - x²-4x-4
y-3= ±√(12-x^2-4x)
y =±√ (12-x^2-4x)+3 

i think that's right, but you get the point. first complete the square to get to standard form, then take the square root and you've got 2 equations to put into calc for one circle.

Last edited by argo4 (2008-05-17 19:09:24)

 

Pages: 1

Board footer

Privacy Policy - © 2025 Jeff Minard