x=-6y^2 + 4y
Use the quadratic formula.
Try completing the square (i.e. as-if it was just a normal quadratic equation of a single variable)... it should work.
okay, so plotting BOTH of these graphs will give me a perpendicular graph of y=-6x^2+4x, correct? each of the functions from the quadratic formula should be each half of a perpendicular parabola.
hhhmmm...I just took the equation and put it like this:
-6y2 + 4y - x = 0
and solved it using the quadratic formula, so
y = ( -b + sqrt(b2 - 4ac) ) / 2a
and the other solution to y is with the negative value of the square root.
I get something like this:
y = 1/3 + sqrt( (2-3x)/18 )
and, of course,
y = 1/3 - sqrt( (2-3x)/18 )
which tells me that x cannot be larger than 2/3, otherwise the radical becomes negative.
So, you can plot y versus x for -inf <= x <= 2/3
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