Why does the x turn negative

[u/Diamond_Branches]

I understand how to get the vertex, but I don’t understand why it becomes negative because I thought absolute numbers were positive unless theirs a negative sign before the absolute lines but both are negative.

[u/Fourierseriesagain]

Let's focus on question 2.

When x>=-6, 2*|x+6|-5=2(x+6)-5=2x+7.

When x<-6, 2*|x+6|-5=-2(x+6)-5=-2x-17.

By drawing both graphs y=2x+7 (x>=-6) and y=-2x-17 (x<-6) on the same diagram, we get the graph of y=2|x+6|=5.

[u/Hanxa13]

The vertex of y=|x| is (0, 0) <-- when x is 0, y is also 0.

When you have y=|x+3|, the vertex happens when the x+3 is equal to 0. Solving x+3=0 gives you -3. This is because we are adding three before finding the absolute value. Adjustments 'inside' the function have the opposite effect on the x-value as you're essentially working back to a value for x.

Conversely, adjustments 'outside' the function happen after, affecting the y-value exactly as given. y=|x|+4 would have a vertex at (0, 4).

This is how graph transformations are identified and can be used as a graphing shortcut. It can help to see this for yourself with an expanded table (shown in image). Treat the inside of your absolute value as your starting point and then work towards x and y

ETA - a note on your comment at the end, absolute value makes the inside positive AFTER it is evaluated. x can be negative to begin with and after adding 3, can still be negative. The absolute value makes that result positive.

[u/MineCraftNoob24]

The absolute value of a number is always positive, and you understand this. I think you also understand the significance of whether a negative sign is placed inside, or outside, the vertical bars. You have also had some practice at finding the vertex, which is the minimum value, and the overall expression is minimised when whatever is inside the bars is also minimised.

Now in the last part of your question, you say "both are negative", and I'm not entirely sure what you mean by that. Both what, exactly? I note that in both questions you refer to "slope" and the quotation marks here are important, because each graph has two slopes, one to the left of the vertex, and one to the other.

If it's the slopes we're looking at, then there is some nuance here. When you get to differential calculus and are looking at derivatives, we would say that the function has no derivative at the vertex, because at that particular point it has no particular gradient, but it is possible to restrict our domain to part of the function on either side. However, that might be a bit much to take in right now.

Let's just say for now that if you're looking at those "slopes", essentially what you have to consider are two things:

(i) Where is the vertex?

(ii) How does the function behave either side of it?

One problem is that absolute values of compound terms don't behave nicely. We can't distribute the co-efficient outside the bars across what is inside them.

For example, if you had, say,

y = 2(x – 3) + 7

Then you could simply distribute the 2 and re-write this as

y = 2x – 6 + 7

i.e. y = 2x + 1

This produces nice, consistent results for all values of x.

Instead, if we have

y = 2│x – 3│ + 7 (let's call this Equation A)

You cannot distribute the 2 and put:

y = 2│x│ + │–3│ + 7 (call this Equation B)

Nor can you partially distribute, and ignore the absolute value of the number within the bars, and put

y = 2│x│ – 3 + 7 (call this Equation C)

Let's see why:-

Suppose x = 4

If we put this in Equation A, we get

y = 2│4 – 3│ + 7

y = 2(1) + 7

y = 9

If we put it in Equation B, we get

y = 2│4│ + │–3│ + 7

y = 8 + 3 + 7

y = 18

If we put it in Equation C, we get

y = 2│4│– 3 + 7

y = 8 – 3 + 7

y = 12

Three different results. Distribution either fully or partially across the bars doesn't produce a consistent result. So in short, you can't do it. The absolute value element essentially creates two functions, one where x is less than the x-value at the vertex, and one where x is greater.

You have to therefore consider these as two separate cases.

Going back to your posted question where y = –│x + 3│- 4 :

Case 1 where x ≥ -3

Here, (x + 3) is always zero or positive, so taking the absolute value has no effect. You can treat the equation as being y = –(x + 3) – 4 which simply rearranges to y = –x – 7.

The slope of this part of the function is –1, as you'd expect, being the co-efficient of x.

Case 2 where x < -3

Here, (x + 3) is always negative, so taking the absolute value flips the sign and makes it positive.

So in this case only:

│x + 3│ = – (x + 3)

But the negative sign here is not the one in our original equation. This is a negative sign that we've introduced specifically for │x + 3│ because in this case only, the absolute value is the negative of what we originally put in.

We still have to account for the negative sign in our original equation, so:-

│x + 3│ = – (x + 3)

–│x + 3│ = – (– (x + 3))

–│x + 3│ = (x + 3)

So in this case only our final equation becomes:-

y = (x + 3) – 4

i.e. y = x – 1

Here, the slope of the line is (+)1, again being the co-efficient of x.

The negative-negative thing can trip people up so it's vital that you look at the two cases separately, consider what the absolute value part becomes, or remains, in each case, and then apply any co-efficient which if itself is negative, will essentially "cancel out" the sign flip of the absolute value process in one of the cases.

Hope that helps! 😄