how would I sketch the curve consisting of the points (x,y) such that e^(4y^2)e^(4y)e^(-x)=e^2

Question

how would I sketch the curve consisting of the points (x,y) such that   e^(4y^2)e^(4y)e^(-x)=e^2

dumbcow · Answer

you can rewrite this as:
$$e^{4y^2 +4y-x} = e^2$$
then take log of both sides
$$4y^2 +4y-x = 2$$
isolate x
$$x = 4y^2 +4y-2$$
this is a sideways parabola
complete the square to get in vertex form
then graph

dumbcow · Answer

$$x = 4(y+\frac{1}{2})^2 -3$$
vertex at (-3, -1/2)

larseighner · Answer

$$\large e^{4y^2}e^{4y}e^{-x}=e^2$$
$$\large e^{4y^2}e^{4y}={e^2 \over e^{-x}} $$
$$\large e^{4y^2}e^{4y}=e^2 e^x $$

Now I argue from the law of exponents, which works with any base, that this implies the equation that @dumbcow found, and the solution proceeds a @dumbcow wrote.