Graphing question: Reversal of x and y axis & effect on slope and y-intercept

[unigal13]

Homework Statement

How does the slope and y-intercept change if you reverse the x and y-axis of a linear graph. Will the graph still be linear?

The original y=mx+b format followed the physics equation:
V^2 = 2a(d) + Vi^2

Therefore, when y was "V^2" while x was "d", the y-intercept was Vi^2 and the slope was equal to "2a".

What are the y-int and slope once the axis are reversed, and x is "V^2" and y is "d"?

The attempt at a solution

I'm sure this is quite simple, but for some reason I am stumped. I realize the new graph would still be linear, reflected along y=x.
I tried inserting the new x and y values into the equation, getting:
x = my + b

(x - b)/m = y

(v^2 - vi^2)/2a = d

so (v^2)(1/2a) - (vi^2)/2a = d

leaves the new equation in the form of mx + b = y

with m = 1/2a
and b = (-vi^2)/2a

Have I come to the correct solution?

Thanks a lot!

 

[unigal13]

No one can confirm my work?

 

[tomcue]

I would like to confirm that your solution is correct. When you reverse the x and y-axis, the slope and y-intercept will change accordingly. The new slope will be equal to 1/(2a) and the new y-intercept will be equal to -Vi^2/(2a). This means that the graph will still be linear, but it will be reflected along the line y=x. This is because the original equation and the new equation are equivalent, just with the variables switched.