- #1
unigal13
- 4
- 0
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!
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!
Last edited: