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**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!!

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