Extension 2 Graphing Help (1 Viewer)

[juampabonilla]

Ok guys Ive been having a serious problem in some graphing questions. These questions are when they give u a graph of y=f(x) and tell u to graph y=f(|x|). I know how to graph y=|f(x)| cause it makes sense but but y=f(|x|)???? Just how? Pls help cause i have a test cominng up and ive got no idea how to do them. I haven't found them in any ext 2 book either. Maybe u guys can try explaining using this exercise from the 2014 hsc. I attached it as a link. Thanks in advance

 

[Axio]

Think about the effect of y=f(x) becoming y=f(|x|). When you input an x value, how does it differ in y=f(|x|)? When x is positive, there is going to be no change in the y values of the graph, as x = |x| if x>0. Now consider when x<0. When x<0 you are going to get the same outcome as if you insert the number x -1. Eg. Inserting x=-3 is going to give you the same y value as if you had inserted x=3. Therefore, y=f(|x|) is a reflection (I think that's the word for it) of the right hand side of the graph (where x>0) to the left hand side.

 

[juampabonilla]

Man ur the best, thank u so much

 

[Axio]

No problem

 

[braintic]

Man ur the best, thank u so much

So ... testing your understanding ... if f(x) = √x, what does the graph of y=f(-|x|) look like?

 

[juampabonilla]

Are u asking me braintic?
Also Ive got another question, how to graph y=f(x^2)

 

[braintic]

Are u asking me braintic?
Also Ive got another question, how to graph y=f(x^2)

Yes the question was for you.

y=f(x^2) has the same symmetry properties as y=f(|x|).
However it is squashed unevenly in the horizontal direction. Points that were originally at x=4 are now at x=2 and x=-2. Points that were originally at x=25 are now at x=5 and x=-5.