calculus
posted by Joe .
Piecewise
f(x) = { (x^2, x<=2) (mx + b, x>2)
I am not sure what the graph would look like for the 2nd function.
Any tips on drawing functions like this would be of great help.

There is no unique graph for y = mx + b without knowing the values of m and b .
The only thing I can surmise is that the straight line will continue on from (2,4), which is the end point of the parabola.
so the endpoint of y = x^2 , x ≤ 2 is (2,4)
dy/dx of the parabola is 2x
at (2,4), dx/dy = 4
so find y = mx + b, with m = 4 and point (2,4)
4 = 4(2) + b
b = 4
so graph y = x^2 up to x = 2, then from there
graph y = 4x  4 , but only for x > 2
(You can use the yintercept of 4 and the point (2,4) to get the straight line, but don't draw it for x < 2 )