Suppose that a point moves along some unknown curve y = f(x) in such a way that at each point (x, y) on the curve the tangent line has a slope . Find an equation for the curve, given that it passes through (1,1).

is just the the derivative of since it it is the slope of the line tangent to the curve at . To find how is defined, we'll just take the anti-derivative of .

Recall:

Solving for :

We still have to solve for . Note that the point has to be on our graph so or .

Solving for :

We can finally write or . Since , we can write it as as well.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.