x^2+y^2=81
I know how to find the intercepts in a regular standard problem, but not when both variables are squared. Any help would be appreciated!
2006-09-03 15:54:18 · 9 answers · asked by linfinity27 2 in Education & Reference ➔ Homework Help
it's called a circle. It's basically a circle centered at the origin.
the interceps for x are : 9 and -9
explanation: when y is 0, then y^2 is also 0.
x^2+0=81
x=9, -9
y: 9 and -9
same goes for y
hope this helps!
2006-09-03 15:56:44 · answer #1 · answered by Anonymous · 1⤊ 0⤋
this is a circle when both terms are squared and added. this also means that the other side of the equation is the radius (of the circle) squared. since the two variables are alone (they dont have anything added or subtracted) the origin is at (0,0). then to find the intercepts you would use the radius, which in this problem is 9. so the x-intercepts would be (-9,0) and (9,0) and the y-intercepts would be (0,-9) and (0,9). hope this helps!
2006-09-03 22:59:42 · answer #2 · answered by leksa27 2 · 1⤊ 0⤋
for example let x=0. Then y^2=81. So y= +/-9. Therefore y intercepts are (0,9) and (0,-9). Then do the same sort of thing for the other axis. (This graph is a circle)
2006-09-03 23:00:21 · answer #3 · answered by banjuja58 4 · 0⤊ 0⤋
actually, it doesn't matter if the variables are squared. you solve like any other problem. plug zero in for x, solve and get x intercept. then plug zero in for y, solve and get y intercept. no different. so what's zero squared? zero. also, don't forget that when you take the square root of 81 you get plus/minus 9. therefore, since this is clearly a circle...
x intercepts = (9,0) and (-9,0)
y intercepts = (0,9) and (0, -9)
2006-09-03 23:02:27 · answer #4 · answered by hmbn 4 · 1⤊ 0⤋
Well, know that when both variables are squared, you get an ellipse (or a circle, which is a special ellipse--in this case it's a circle) (this is with addition) or a hyperbola (this is when one variable is subtracted from the other).
2006-09-03 23:47:38 · answer #5 · answered by Anonymous · 0⤊ 0⤋
try solving it where you find the x + y= the square root of 81 because to similify you would find the square root of the whole equation so the square root of x^2 is x and the same with y, you get x+y=the square root of 81
2006-09-03 23:30:28 · answer #6 · answered by Anonymous · 0⤊ 0⤋
your equation is the equation of a circle.
The centre of the circle would be (0,0)
whereas the radius of the circle would be the square root of 81, which is 9
2006-09-03 23:34:20 · answer #7 · answered by e.z p.z 2 · 0⤊ 0⤋
x-intercepts:
x^2 + 0 = 81
x = +/- 9
(-9,0) and (9,0)
y-intercepts
0 + y^2 = 81
y = +/-9
(0,-9) and (0,9)
2006-09-03 22:56:12 · answer #8 · answered by Anonymous · 2⤊ 0⤋
It's a circle :)
2006-09-03 22:55:53 · answer #9 · answered by Anonymous · 1⤊ 1⤋