not clear what you mean by solve for x and y,
1. x^2=4y+13
is a parabola, which can be arranged as y = (1/4)x^2 - 13/4
or x = ±√(4y + 13)
2. (x-.56)^2/.4 + (y+.39)^2/1.6 = .5
an ellipse centre at (.56, -.39) , with a^2 = .2 and b^2 = .8
3. x^2+4x-2.52 , not even an equation, if it is equal to y, you have another parabola
4. (y+.8)^2/.2 - (x+0)^2/1 = 1 <---- a hyperbola
Solev for x and y:
- x^2=4y+13
- ((x-.56)^2)/.4+((y+.39)^2)/1.6=.5
- x^2+4x-2.52
- ((y+.8)^2)/.2-((x+0)^2)/1=1
Can someone solve these because i wasnt able to find the answers for like an hour:(
2 answers
If you want the x- and y-intercepts, just plug in y=0 or x=0 and solve for x or y.
1 answer