Question

help!

Answer 1

@amistre64

Answer 2

essentially, an inverse function swaps xs and ys

Answer 3

let y = f(x); and solve x = f(y) for y

Answer 4

might want to "complete the square" first tho

Answer 5

\[x = y^2 + yx\]
\[x+\frac{b^2}{4} = x^2 + bx+\frac{b^2}{4}\]
\[x+\frac{b^2}{4} = (y+\frac b2)^2\]
\[\sqrt{x+\frac{b^2}{4}} = y+\frac b2\]
\[\sqrt{x+\frac{b^2}{4}} -\frac b2= y\]

Answer 6

\[x=\frac{ -9+\sqrt{81+4y} }{ 2 }\]

Answer 7

just switch x and y

Answer 8

@amistre64 yeah i figured you would have to square but was not sure. thanks!

Answer 9

my latex got a little messed up :)

Answer 10

DM went ahead and solved for "x", which is fine as well