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C3 help!!!!

for graphing y=f(IxI)
why is it a reflection in the y axis, i thought the makes all X values positive and that means Y values too????
Original post by JustgotAstar
for graphing y=f(IxI)
why is it a reflection in the y axis, i thought the makes all X values positive and that means Y values too????

The y value of the negative x values becomes positive. By drawing this part of the graph with the positive parts, it looks like it's a reflection in the y axis.
It's just a linear function that meets the x axis at y=0 so the reflection is in y=0, the y axis.

Spoiler

(edited 6 years ago)
If you take f(x) = x that will mean y = |x|
So the y value when x = -2 will be the same as when x = 2 so it is going to be reflected in the y axis

However if f(x) = x-4 that will mean y = |x|-4 so values where |x|<4 the y value will be negative, so it doesn't always follow that y = f(|x|) produces positive values of y, it depends of the function f(x)
The X value can be a negative, however the modulus of a plus or minus value is always positive i.e so when X = -3, Y = 3 as the -3 is converted into a 3, so the Y value never goes below 0.
Original post by JustgotAstar
for graphing y=f(IxI)
why is it a reflection in the y axis, i thought the makes all X values positive and that means Y values too????


Try plotting the simplest function: y = x and then plot y = modulus x. See what happens!
The value of y will be positive regardless of |x| being positive or negative because the modulus function states that the value will be positive hence the positive y values. I just remember it like this because it makes it easier to understand and apply unless you are doing complicated problems.
Original post by JustgotAstar
for graphing y=f(IxI)
why is it a reflection in the y axis, i thought the makes all X values positive and that means Y values too????


For a function f(x)f(x), replacing xxx \mapsto |x| means that f(5)=f(5)f(|-5|)=f(|5|) and f(1)=f(1)f(|-1|)=f(|1|), and so on until x=0x=0. In fact, f(x)=f(x)f(|-x|)=f(|x|). This is symmetrical around the line x=0x=0, which is the y-axis and hence the x>0x>0 portion looks reflected in the y-axis to look like a reflection on x<0x<0 portion of the graph.

For all OUTPUTS to be positive, ie your Y values as you'd refer to them, you'd need to have f(x)|f(x)| or something like that with the modulus brackets on the entire function.

Simple example, take f(x)=(x1)(x2)=x23x+2f(x)=(x-1)(x-2)=x^2-3x+2 which takes strictly negative values for 1<x<21<x<2, then graphing f(x)=x23x+2f(|x|)=|x|^2-3|x|+2 shows you the reflection of the x>0x>0 portion of f(x)f(x) in the y-axis, and so the graph must dip again below 0 in the region 2<x<1-2<x<-1
(edited 6 years ago)
Wow, thanks guys, you all made it really clear particularly RDKGames

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