Question 1

In example problem 1, why isn’t the graph shifted 1 unit to the left instead of to the right?

Accepted Answer

It is shifted to the right because `x-1` would make it `0` when `x=1` because
 `x=1`
`1-1=0`
So, we always want the absolute value part of the equation to be equal to 0 when we use x as the horizontal shifting.

While, the vertical part goes *up with + not down* because when,
`y=a∣x−h∣+k`
`y-k=a|x-h|`
So basically we transpose it to make it easier to distinguish.

Question 2

is there any easier steps to explain this type of lesson

Accepted Answer

Maybe I can better explain

when you have an absolute value function you want to look at what are in the places of a, h and k.  a|x-h|+k  Specifically you want to look at h and k first.  Normally the tip of the V shape is at (0,0) this changes depending on h and k.  specifically it moves the tip to (h,k) so if you have |x+5|-7 then the tip of the V shape goes to (-5,-7).  if you wonder why it is -5 even though we are adding 5, you just need to look at the original a|x-h|+k  if we had -5 then it would be just like that, but since it is +5, we have to look at it as - -5, minus negative 5.  so if it helps, the x coordinate is kinda backwards.

After the V tip you then look at a.  treat it like a linear equation where a is the slope.  so if a was -3 that's down 3 right 1 using rise over run.  then, since it's an absolute value function  you need to know that the same line goesalong the left to make that V shape, so -5 would mean on the left down 3 and left 1.

if you ever have something like a|bx-h|+k where there is a number in front of the x you need to get rid of it  if you are not aware of factoring  this is what it would look like a|b||x - h/b|+k where a|b| becomes the new "a" and h/b becomes the new h, then you would solve it normally.  The point being you always want x by itself for this.  Also, keep in mind that even if inside the absolute value bars if b was negative, outside it becomes positive.

Let me know if that didn't help, or if there is a specific function you are struggling with, or maybe would even like some to try out.

Question 3

So is h is positive that means that it is actually negative? Is that why if its x + 3 on the graph you go to negative 3?

Accepted Answer

No, if it is positive it means I move in the negative direction, but if h is negative I move in the positive direction, it does not change the sign of h.  The idea is that what value of x would make the inside of the absolute value (or other function) 0, so if you have x + 3, it would require x = -3 to be 0, thus causing a shift in the negative direction.  The other idea is that since the formula has | x - h | + k where (h,k) is the vertex, then using x+ 3 would actually be x - (- 3) so -3 would cause a shift to the left.

Question 4

What if there's something in front of the x? In my homework, there's a problem that says a(x) = |5x|. How would I graph that?

Accepted Answer

You would simplify the expression inside the `| |` and take the absolute value of the result.  For example
|5*0| = |0| = 0
|5*1| = |5| = 5
|5*-1| = |-5| = 5
|5*2| = |10| = 10
|5*-2| = |-10| = 10

Question 5

I am confused on how you know if the vertex is a minimum or maximum point.

Accepted Answer

The general form of the absolute value function is:
f(x) = a|x-h|+k
When "a" is negative, the V-shape graph opens downward and the vertex is the maximum.
When "a" is positive, the V-shape graph opens upward and the vertex is a minimum.
Hope this helps.

Question 6

I just need to know how to graph absolute value with a table!

Accepted Answer

To graph the absolute value, you need to know where the vertex is, so the numbers will be going down, then turn and start going back up, or going up and then turn and start going down, the middle point will be the vertex and should be the center point of the table.  if you learned vertex form in quadratics, it is the same as in absolute value, so vertex form gives f(x) = a | x - h | + k has a vertex at (h,k), example:  F(x) = | x - 4 | - 3 has a vertex at (4,3).

Question 7

This article says that the general form of absolute value equations is f (x) = a |x - h| + k.
I have the problem y = 4|x|.
How do I solve it when I have a different form?
 I would really appreciate help.

Accepted Answer

This is in the form identified in the article.  h and k both equal zero so the vertex is the point (h,k)=(0,0).  a=4 so the graph opens up and is steep.

Question 8

I have a table where the coordinates are in a table, the coordinates are
(4,7) (5,6) (6,5) (7,4) (8,5) 
it is wanting me to find the domain which I found, but I don't know how to find the range, intercepts, vertex, or the max or min.
How do I find the range, intercepts, vertex, and the max or min?

Accepted Answer

The range refers to the possible y values.  Based on the table you gave us, it looks like there is a minimum y value.  It looks like the y values start at 7, go down, and then start going up again.  The lowest y value is going to be the turning point of the variable.  The turning point is always going to be the minimum or the maximum.  The turning point is your vertex.

Since there appears to be a lowest y value, the graph probably doesn't have a highest y value.  It depends a little on the question.

The vertex is where the graph turns.  It should happen at the lowest y value (or the highest, if that makes more sense for the problem.  If the graph goes up forever, then it should be the lowest y value).  The vertex is also where the min, or max of the function is... but remember, whatever the vertex is (max or min), the other one (min or max) is probably infinity.

For intercepts... based on the table, does it look like it ever crosses the x axis?  Use the slope of the line (before it turns, or after it turns), to figure out where the line goes beyond the points that it gave you.  For the Y-intercept, what is the Y value when X is 0?

Get ready for Algebra 2

Unit 3: Lesson 7

Absolute value graphs review

Example problem 1

Example problem 2

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