Let's think a little bit about the graphs of absolute value
functions.
And I've defined one right over here. f
of x is equal to negative 3 times the absolute value of x
minus 1 plus 9.
And then we've constrained its domain.
This is for x being-- or negative 4
is less than or equal to x, which is less than
or equal to 5.
And I encourage you to pause this video
and try to graph this on your own.
And clearly you could graph this by just trying out
different x-values and plotting the corresponding x and f
of x values, but really think about the structure
of this expression to think about
where does it hit a maximum or minimum point?
And where would the vertex of it actually be?
So let's look at this just holistically
before we even try some points.
So we have an absolute value.
This absolute value here is clearly
going to always be non-negative, and then we're
multiplying it by a negative number.
So this entire thing, you have a negative times
a non-negative, so this whole thing is always
going to be non-positive.
The reason why I say non-positive instead
of negative is that because it could be 0.
So this whole part of the expression
is either going to be 0, or it is
going to be a negative number.
Now, if this is always non-positive,
what is the maximum value of this function?
Well, the maximum value happens when
the absolute value is equal to 0.
When the absolute value part of the expression is equal to 0,
then this part is 0, and the function hits its maximum value
at f of x is equal to 9.
So when does our absolute value part of this expression
equal 0?
Well, let's think about it.
When does the absolute value of x minus 1
is going to be equal to 0 if x minus 1 is equal to 0.
If x minus 1 is equal to 0-- or if you add 1 to both sides--
you get x is equal to 1.
So when x equals 1, we hit our maximum point.
This thing becomes 0, which makes
this whole thing equal to 0.
And so we get the point 1 comma 9.
So let's graph that point.
And actually, let me do it in a color that's
easier to see on a white background.
1 comma 9 is the maximum point right over here.
And you could also view this as the vertex
of this absolute value function.
So 1 comma 9 is that point right over there.
And so we can already guess the general shape of this.
An absolute value function either
looks something like that, or it looks something like that.
It would look like this if the coefficient right over here--
or i guess maybe I shouldn't call it a coefficient.
If the thing multiplying the absolute value was positive,
then would it be upward opening like that.
But since it is negative, it is going to be downward opening.
We hit a maximum point at the vertex in the case
we're just talking about for the reasons we just discussed.
But then what happens there?
What happens to the right of that vertex,
and what happens to the left of this vertex?
We don't have to do it this way.
Once again, you could just try out points if you like,
but I like to really analyze what's going on here.
So let's break up this function definition into two pieces.
So let's say that f of x is equal to--
and we're going to break it up into two pieces, one
piece to the right of the vertex and one
piece to the left of the vertex.
So to the right of the vertex, that's
going to be 1 is less than or equal to x, which is less than
or equal to 5.
And to the left of the vertex-- so
this is to the right of the vertex.
To the left of the vertex, that's going to be-- let
me do this in a different color.
That's going to be negative 4 is less than or equal to x,
and I'll just say less than 1.
So just to be clear, our whole domain
was between negative 4 and 5.
Between negative 4 and 5 was our entire domain.
Now, I'm breaking it up to the right of the vertex
and the left of the vertex.
So this right over here is to the right of the vertex,
and this right over here is to the left of the vertex.
So to the right of the vertex, for x
being greater than or equal to 1, this absolute value-- or one
way to think about it, when x is greater than or equal to 1,
x minus 1 is going to be positive,
and the absolute value of a positive thing
is just going to be that same value.
So for x is greater than or equal to 1,
our function is going to be negative 3 times x minus 1 plus
9.
Or if you wanted to distribute the negative 3,
this would be the same thing.
Actually, let me write it over here so that we
can write the simplified version.
So it would be-- let me give myself some space.
So for x is greater than or equal to 1,
we don't have to think about the absolute value.
So it would be negative 3 times x minus 1
plus 9, which is negative 3x plus 3 plus 9,
or negative 3x plus 12.
So it's going to be negative 3x plus 12
when x is greater than or equal to 1.
And the really interesting thing, the slope
here is negative.
We have a slope of negative 3.
So starting from this point, if we have a slope of negative 3,
that means every time we move 1, increase in the x-direction,
we're going to decrease by 3 in the y-direction,
increase by x, decrease by y by 3.
So it's a slope of negative 3.
So it will look like that.
And let me actually do it in purple
so that you can actually see it.
That's not purple.
There you go.
That's purple.
All right.
So it would look like that.
And actually let me just make it clear.
This is what I'm talking about right over here.
That piece of the function is going to look like that.
Now, what about when x is less than 1?
Well, when x is less than 1, x minus 1
is going to be negative, and so the absolute value
is going to be the negative of that actual function.
So when x is less than 1-- let me write this.
The absolute value of x minus 1 is
equal to the negative of x minus 1, which is the same thing
as 1 minus x.
So our function is going to be negative 3 times 1 minus x
plus 9, which is the same thing as-- let's
see, if you distribute the negative 3, it's negative 3
plus 3x plus 9, which is equal to 3x plus 6.
So the important thing to realize here
is that our slope is positive 3.
And you could even say that we have a y-intercept here at 6,
so it would be this point right over here,
and the line would look like this.
When your slope is positive 3, you increase x by 1,
you increase y by 3.
Or another way of thinking about it, if you decrease x by 1,
then you decrease y by 3.
So it would look like that.
And we are done.
We have graphed this absolute value equation.
Now, the whole reason why I did this
is just so we learn it a little bit deeper.
But there would have been a faster way of doing this.
You could have just said, hey, this
is going to hit a maximum point when x is equal to 1
because this non-positive thing is going to hit 0.
So you could have said 1 comma 9 is the vertex.
And then you could have just tried out the two endpoints,
knowing that between the endpoints and the vertex,
you're essentially just dealing with a line.
So you could have just evaluated f of negative 4 and f of 5.
You could have figured out these endpoints and drawn the lines.
But now we understand it deeper, and that's always
a little bit more exciting.