Question

find the inverse of y=2x^2-5

Answer 1

Notice that for the inverse, the x and y switch their places. So first replace y with x and x with y, then re-arrange and solve for y.

1) Switch x and y:\[\bf y=2x^2-5 \rightarrow x = 2y^2-5\]2) Solve for y:\[\bf x=2y^2-5 \rightarrow \frac{ x+5 }{ 2 }=y^2 \implies y^2 = \frac{x+5 }{ 2 }\]
@selenamalter

Answer 2

Alternatively, you could take the square root of both sides but then since square is always positive, you will need to have it as:\[\bf y=\pm \sqrt{\frac{ x+5 }{ 2 }}\]The "plus-minus" is necessary since we need two parts, one above the x-axis and one below the axis. Leaving the square root without the "pm" sign will only result in the positive part. To get rid of this confusion, it's best to leave the function in terms of "y^2" as I did in the previous reply since you don't have to deal with the "pm" sign.

@selenamalter