Question 1

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 2

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 3

How would we utilize this in real life? For what careers?

Accepted Answer

mathematician

Question 4

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 5

How do you identify the vertex y intercept and x intercept

Accepted Answer

Hey there,

I'm not an expert here, but it was an interesting exercise to figure out the answer to your questions and I figured I might as well post it here. Sorry if it's too much of a wall of text to get through.

Just to recapitulate, the general form is:
f(x) = a|x−h| + k

The vertex is located at point (h,k). The minimum or maximum (depending on whether a is positive or negative) of the graph is at the point where x - h = 0. This is the same as saying x = h, which gives us the x-coordinate of the vertex. As for the y-coordinate: since we just saw that |x-h| = 0, a|x−h| must also be 0, which only leaves us with k.

To find the y-intercept, we can set x to 0. In the general formula, that means:

f(0) = a|0−h| + k
f(0) = a|h| + k

Which gives us, as a general rule, (0,(a|h|+k)) as the y-intercept. Taking one of the examples, f(x)= |x−1| + 5 where a=1, h=1 and k=5: the y-coordinate of the intercept is 1|1| + 5 = 6, which means the intercept is at (0,6).

The method I thought of to find the x-intercepts a bit more involved, maybe someone else knows an easier way. I basically just used algebra.

There can be 0 or 2 x-intercepts depending on the value of k and a.
There will be two x-intercepts if:
k > 0 and a < 0
or
k < 0 and a > 0
and no x-intercepts otherwise. That said, let's use the general form again and set the result of the function to 0 and try to solve for x.

0 = a|x−h| + k
-k = a|x−h|
-k/a = |x-h|
|-k/a| = x-h
|-k/a| + h = x
Here it gets a bit tricky. There can be two possible values such that their absolute value together with h adds up to x: -k/a and k/a, since both evaluate to the same absolute value. But since we're looking for two intercepts, it actually makes sense that there are two possible results for x:
-k/a + h = x and k/a + h = x

I tested this with f(x)= -2|x+5| +4
According to my result, -(4/-2) +(-5) and 4/-2 +(-5) should be the x-coordinates of this graph's y-intersects: -3 and -7, and it checks out! It kind of makes sense as well: we're dividing k - the difference in y from the x-axis at the maximum - by the slope. This should give as result the difference in x from the maximum, and then we're adding the amount by which the maximum was shifted.

Cheers if someone actually read all of these words.

Question 6

what if x has a coefficient

Accepted Answer

Great Question. When x has a non-one and non-zero coefficient, the curve stretches or shrinks.
When coefficient of x is larger than one, then the curve shrinks along the x-axis with the scale of 1 / (coefficient).
When coefficient of x is smaller than one but larger than zero, the curve expands with the scale of 1 / (coefficient).
If it's negative then it's a reflection.

Question 7

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 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?

Question 9

Can Someone solve this hard inequality I've been stuck in for over 30 years solving it.
|x+6|>|x-6|

Accepted Answer

𝑥 + 6 = 0 ⇒ 𝑥 = −6
𝑥 − 6 = 0 ⇒ 𝑥 = 6

This means that there are 3 scenarios we need to analyze:
𝑥 < −6
−6 ≤ 𝑥 < 6
𝑥 ≥ 6

– – – Scenario 1 – – –

𝑥 < −6
⇒
|𝑥 + 6| = −(𝑥 + 6) = −𝑥 − 6
|𝑥 − 6| = −(𝑥 − 6) = 6 − 𝑥

Thus, |𝑥 + 6| > |𝑥 − 6|
⇒ −𝑥 − 6 > 6 − 𝑥
⇒ −6 > 6, which is not true.

This means that the inequality |𝑥 + 6| > |𝑥 − 6|
is not true for any 𝑥 < −6.

– – – Scenario 2 – – –

−6 ≤ 𝑥 < 6
⇒
|𝑥 + 6| = 𝑥 + 6
|𝑥 − 6| = −(𝑥 − 6) = 6 − 𝑥

Thus, |𝑥 + 6| > |𝑥 − 6|
⇒ 𝑥 + 6 > 6 − 𝑥
⇒ 𝑥 > −𝑥
⇒ 2𝑥 > 0
⇒ 𝑥 > 0

Thereby |𝑥 + 6| > |𝑥 − 6|
is true when −6 ≤ 𝑥 < 6 AND 𝑥 > 0,
i.e., when 0 < 𝑥 < 6

– – – Scenario 3 – – –

𝑥 ≥ 6
⇒
|𝑥 + 6| = 𝑥 + 6
|𝑥 − 6| = 𝑥 − 6

Thus, |𝑥 + 6| > |𝑥 − 6|
⇒ 𝑥 + 6 > 𝑥 − 6
⇒ 6 > −6, which is true.

Thereby |𝑥 + 6| > |𝑥 − 6|
is true for all 𝑥 ≥ 6

– – –

In conclusion, |𝑥 + 6| > |𝑥 − 6|
is true when 0 < 𝑥 < 6 OR 𝑥 ≥ 6,
i.e., for 𝑥 > 0.

In other words,
𝑥 > 0 is the solution to |𝑥 + 6| > |𝑥 − 6|

Get ready for Algebra 2

Course: Get ready for Algebra 2 > Unit 3

Absolute value graphs review

Example problem 1

Example problem 2

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