Look at the graph here.
How would you draw the following graphs;
Given f(x) = |x|
a) y = 2 - f(x-1)
Would it be the original modulus graph turned upside down (like a sharp n) and the top is point (1,2)?
b) y = f(1-x) -1
Not too sure on this one :/
Cheers
Yes. When x= 1, f(x- 1)= f(0)= |0|= 0 for y= 2- 0= 2- that is the point (1, 2). If x> 1, x- 1> 0 so f(x- 1)= |x- 1|= x- 1 and 2- f(x- 1)= 2- (x- 1)= 3- x. That is a line through (1, 2) with slope negative 1. With x= 0, x- 1< 0 so f(x- 1)= |x- 1|= -(x- 1)= 1- x and 2- f(x- 1)= 2- (1- x)= 1+ x. That is a line through (1, 2) with slope 1.
When x= 1, 1- x= 0 so f(1- x)= f(0)= 0. This the point (1, 0). If x> 1, 1- x< 0 so f(1- x)= -(1- x)= x- 1. That is a line through (1, 0) with slope 1.b) y = f(1-x) -1
If x< 1, 1- x> 0 so f(1- x)= |1- x|= 1- x. That is a line through (1, 0) with slope -1. It is just |x| shifted 1 place to the right.
Not too sure on this one :/
Cheers