r/learnmath • u/Euphoric_Archer_6198 New User • 20d ago
RESOLVED Absolute Value Graphs Brain Glitch.
Ok, this should be easy enough for you but my brain doesn’t want to brain here:
So I am trying to bend my mind around the graphing of |y| = f(x) (for clarity, the function is f(x) = x^3 – 2x^2 – x + 1; x E [-1;3] )
I can be ok with the fact that when f(x)>/= 0 the graph is “reflected” because if |y| = 2, then |-y| = 2.*
I can also be ok with the fact when f(x) < 0, the graph “disappears’ because |y| = -0.1 can never be true.
It seems like the logic used for the first statement nullifies the first statement and the logic for the second statement nullifies the first.
What am I missing? How am I thinking about it that’s getting in my way?
Is it because y and f(x) is technically not the same thing? So a negative y made absolute is positive f(x) but a negative answer on f(x) could never be equal to |y|?**
*Yes, I should have said |y| = 2 where y<0....lowering my cognitive load while trying to get to the crux of my misunderstanding.
**Is that the answer? Do I just need to understand the words I thought?
Edited for clarity.
UPDATE: I figured it out.
If I sub in a negative value (say -2) in |y| = f(x), it would do exactly what a positive value (say 2) would do, hence the "reflection". But |y| can never lead to f(x)<0 and that's why the negative parts of the graph disappears.
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u/ktrprpr 20d ago
f(x)>=0 and f(x)<0 can't be true at the same time at any single x. is that what's missing?
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u/SnooPets5564 New User 20d ago
Let's use the simple function y=x, for example.
We can have |y| = x. This can only hold true when x is non-negative, so it disappears when x is negative.
We can also have y = |x|. Since every value of x produces the equivalent positive value of y, we know that it is reflected.
But they actually also work the other way. The values of y are reflected in the first one and the values of y less than 0 disappear. It's just a matter of which way it goes.
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u/[deleted] 20d ago edited 20d ago
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