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.

calculus 
Reiny
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 )
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