The general form of an absolute value function is f(x)=a|x-h|+k. From this form, we can draw graphs. This article reviews how to draw the graphs of absolute value functions.
Log in torresmc8760 4 years agoPosted 4 years ago. Direct link to torresmc8760's post “is there any easier steps...” is there any easier steps to explain this type of lesson • (21 votes) loumast17 4 years agoPosted 4 years ago. Direct link to loumast17's post “Maybe I can better explai...” 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. (79 votes) Riad Hajiyev a year agoPosted a year ago. Direct link to Riad Hajiyev's post “If someone needs:Horizon...” If someone needs: • (20 votes) TheAvenger1621 7 years agoPosted 7 years ago. Direct link to TheAvenger1621's post “In example problem 1, why...” In example problem 1, why isn’t the graph shifted 1 unit to the left instead of to the right? • (7 votes) Karmanyaah Malhotra 6 years agoPosted 6 years ago. Direct link to Karmanyaah Malhotra's post “It is shifted to the righ...” It is shifted to the right because While, the vertical part goes up with + not down because when, (23 votes) Chrysopraze a year agoPosted a year ago. Direct link to Chrysopraze's post “How would we utilize this...” How would we utilize this in real life? For what careers? • (5 votes) JitterDew a year agoPosted a year ago. Direct link to JitterDew's post “mathematician” mathematician (18 votes) NNAMNO EKPUNOBI 9 months agoPosted 9 months ago. Direct link to NNAMNO EKPUNOBI's post “How do you identify the v...” How do you identify the vertex y intercept and x intercept • (1 vote) jan.f.miller 9 months agoPosted 9 months ago. Direct link to jan.f.miller's post “Hey there,I'm not an ex...” 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: 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 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. 0 = a|x−h| + k I tested this with f(x)= -2|x+5| +4 Cheers if someone actually read all of these words. (13 votes) jhuang 6 years agoPosted 6 years ago. Direct link to jhuang's post “So is h is positive that ...” 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? • (2 votes) David Severin 6 years agoPosted 6 years ago. Direct link to David Severin's post “No, if it is positive it ...” 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. (7 votes) Julicz 2 years agoPosted 2 years ago. Direct link to Julicz's post “Can Someone solve this ha...” Can Someone solve this hard inequality I've been stuck in for over 30 years solving it. • (0 votes) Jerry Nilsson 2 years agoPosted 2 years ago. Direct link to Jerry Nilsson's post “𝑥 + 6 = 0 ⇒ 𝑥 = −6𝑥 −...” 𝑥 + 6 = 0 ⇒ 𝑥 = −6 This means that there are 3 scenarios we need to analyze: – – – Scenario 1 – – – 𝑥 < −6 Thus, |𝑥 + 6| > |𝑥 − 6| This means that the inequality |𝑥 + 6| > |𝑥 − 6| – – – Scenario 2 – – – −6 ≤ 𝑥 < 6 Thus, |𝑥 + 6| > |𝑥 − 6| Thereby |𝑥 + 6| > |𝑥 − 6| – – – Scenario 3 – – – 𝑥 ≥ 6 Thus, |𝑥 + 6| > |𝑥 − 6| Thereby |𝑥 + 6| > |𝑥 − 6| – – – In conclusion, |𝑥 + 6| > |𝑥 − 6| In other words, (10 votes) TAYLOR191 5 years agoPosted 5 years ago. Direct link to TAYLOR191's post “I am confused on how you ...” I am confused on how you know if the vertex is a minimum or maximum point. • (0 votes) Kim Seidel 5 years agoPosted 5 years ago. Direct link to Kim Seidel's post “The general form of the a...” The general form of the absolute value function is: (10 votes) Rios, Sofia 10 months agoPosted 10 months ago. Direct link to Rios, Sofia's post “what if x has a coefficie...” what if x has a coefficient • (1 vote) joshua 10 months agoPosted 10 months ago. Direct link to joshua's post “Great Question. When x ha...” Great Question. When x has a non-one and non-zero coefficient, the curve stretches or shrinks. (7 votes) dannni2019 7 years agoPosted 7 years ago. Direct link to dannni2019's post “I have a table where the ...” I have a table where the coordinates are in a table, the coordinates are • (0 votes) VincentTheFrugal 7 years agoPosted 7 years ago. Direct link to VincentTheFrugal's post “The range refers to the p...” 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? (6 votes)Want to join the conversation?
Horizontal shift : y = f(x+b)
Vertical shift: y = f(x) +d
Reflection about the X-axis : y = -f(x)
Reflection about the Y-axis : y = f(-x)
Stretch/Compress in the X direction: y = f(a * x)
Stretch/Compress in the Y direction: y=f(x) * ax-1 would make it 0 when x=1 because
x=11-1=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.y=a∣x−h∣+ky-k=a|x-h|
So basically we transpose it to make it easier to distinguish.
f(x) = a|x−h| + k
f(0) = a|h| + k
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.
-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
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.
|x+6|>|x-6|
𝑥 − 6 = 0 ⇒ 𝑥 = 6
𝑥 < −6
−6 ≤ 𝑥 < 6
𝑥 ≥ 6
⇒
|𝑥 + 6| = −(𝑥 + 6) = −𝑥 − 6
|𝑥 − 6| = −(𝑥 − 6) = 6 − 𝑥
⇒ −𝑥 − 6 > 6 − 𝑥
⇒ −6 > 6, which is not true.
is not true for any 𝑥 < −6.
⇒
|𝑥 + 6| = 𝑥 + 6
|𝑥 − 6| = −(𝑥 − 6) = 6 − 𝑥
⇒ 𝑥 + 6 > 6 − 𝑥
⇒ 𝑥 > −𝑥
⇒ 2𝑥 > 0
⇒ 𝑥 > 0
is true when −6 ≤ 𝑥 < 6 AND 𝑥 > 0,
i.e., when 0 < 𝑥 < 6
⇒
|𝑥 + 6| = 𝑥 + 6
|𝑥 − 6| = 𝑥 − 6
⇒ 𝑥 + 6 > 𝑥 − 6
⇒ 6 > −6, which is true.
is true for all 𝑥 ≥ 6
is true when 0 < 𝑥 < 6 OR 𝑥 ≥ 6,
i.e., for 𝑥 > 0.
𝑥 > 0 is the solution to |𝑥 + 6| > |𝑥 − 6|
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.
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.
(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?
