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September 21, 2014

September 21, 2014

Posted by **Joe** on Sunday, October 2, 2011 at 4:37am.

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**, Sunday, October 2, 2011 at 8:58amThere 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 y-intercept of -4 and the point (2,4) to get the straight line, but don't draw it for x < 2 )

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