I'm going to get the same function value at the negative value.

I can get to any negative value if I make x negative enough.

Absolute value of negative 3 is positive 3.

Absolute value of negative 3 is positive 3.

But then take the negative value of it.

Since there's 5 times the absolute value of negative 2 minus 3, the absolute value of negative 2 is positive 2.

Absolute value of negative 8 is 8 and absolute value of 8 is 8.

Or by invoking square root with a negative value?

But then for negative values, the absolute value of x is equal to negative x.

If the step is negative, the start value must be greater then the end value.

So its x value is going to be negative 4.

Because negative 2 has an absolute value less than 3.

We have a negative x value, our x value is negative, so we're gonna move to the left, we're gonna be on this side.

If pixels are negative, we just ignore the negative sign and map back to the absolute value.

Returns a positive value if a b, zero if a b and negative value if a b.

And then finally, we have the absolute value of negative 3.

The absolute value of negative 11, only 11 away from 0.

So this is equal to the absolute value of negative 4.

So it's the same thing as the absolute value of negative 10.

5 times the absolute value of negative 5 plus 3 minus 3.

But if it gets too far negative, if it goes to negative 3, we're cool, negative 4, negative 5, negative 6, negative 7, we're still cool, but then at negative 8, all of a sudden the absolute value isn't going to be less than this.

If it's this diagonal, which has a higher value, then it becomes negative.

If I had a negative value, it would be going to the left.

If this was any other non negative value, you would have two values.

So for every other value other than negative 1, the value of this, of f of x, is equal to 2.

Let's go negative, negative 1, negative 2, negative 3, negative 4, negative 5, negative 6.

But how to figure out the absolute value of that negative number, it's distance.

Negative 2, take the absolute value, you get 2 times 1 2 is 1.

This is implicitly saying that the absolute value of y has to be negative.

We have 5 times the absolute value of negative 1 plus 3 minus 3.

Negative. Negative.

Negative 10, negative 9, negative 8, negative 7.

So, now we have a quadratic that points down into the negative space whose maximum value is 0 and otherwise, it's strictly negative.

I'm going to subtract 12 from it, but I could get to any negative value.

So you'd say that the absolute value of negative 3 is equal to positive 3.

So the absolute value of 8 minus 12, which is negative 4, is positive 4.

Negative 5. Negative 6. Negative 7.

Uh, negative, negative!

Negative. Negative. Jason?

So let's put y as negative 1, so we get negative 1 squared minus 2 times negative 1, that's the value of y, is equal to when x is equal to 0.

Because the way our domain was constrained, this value right here is going to be negative.

So we get the absolute value of y is less than or equal to negative 8.5.

You know, you take the absolute value of negative 6, that's only 6 away from 0.

If you take the absolute value of negative 30, it's going to be greater than 21.

less than negative 5 7, its absolute value is going to be greater than 5 7.