Thread: Parabola?

I used three different equations to generate these graphs?

Can anyone work out what i could have typed?

Since you have no coordinate system shown, it is impossible to say what formulas you might have used.

Ok. i should explain. I have messed with the scaling to make them look identical.
What equations could draw 'parabola-like' other than the standard quadratic?

Any parabola with vertical axis is of the form $y= a(x- x_0)^+ y_0$. A parabola with horizontal axis is of the form $x= a(y- y_0)^2+ x_0$. A parabola with axis that is neither horizontal nor vertical can be derived from those through a rotation.

Any conic section (parabola, ellipse, circle, hyperbola) has equation of the form $Ax^2+ Bxy+ Cy^2+ Dx+ Ey+ F= 0$. The conic section is a parabola if and only if $B^2- 4AC= 0$

(Except for "degenerate cases" which reduce to a single line or two parallel lines. For example if A= C= 1, B= 2, D= E= F= 0 that formula is $x^2+ 2xy+ y^2= (x+ y)^2= 0$ which is true for y= -x, a single line.If A= C= 1, B= 2. D= E= 1, F= 0, the equation is $x^2+2xy+ y^2+ z+ y= (x+ y)^2+ (x+ y)= (x+ y)(x+ y+ 1)= 0$, the parallel lines y= -x an y= -x- 1.)

Thanks. I am getting CODECOGS equation quota exceeded so i cant really see what you have typed although i can work some of it out.
The three equations i typed in were y=x^2, y=coshx, and y^2-x^2=1 ( ignoring one section of the graph)

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