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

Its hard to remember to go left or right.

Accepted Answer

Try to remember this: it's counter-intuitive. If it's -x, the graph is shifted right. if it's positive x, it's shifted to the left. Hope this helps you remember better!

Question 6

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 7

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 8

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|

Question 9

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.

Course: Integrated math 2 > Unit 1

Absolute value graphs review

Example problem 1

Example problem 2

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