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

Printable View

- Sep 16th 2013, 01:57 PMadamcobzModulus graph transformations
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 - Sep 16th 2013, 02:10 PMPlatoRe: Modulus graph transformations
Look at the graph here.

- Sep 16th 2013, 02:21 PMHallsofIvyRe: Modulus graph transformations
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.

Quote:

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.

Quote:

Not too sure on this one :/

Cheers