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Thread: Sketch the Graph of an Absolute Value Function, and Apply Transforations of the form.

  1. #1
    Master Sorcerer Lopyhupis's Avatar
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    Default Sketch the Graph of an Absolute Value Function, and Apply Transforations of the form.

    Help,

    y=a*|1/b(x-h7)|+k

    I need help, someone explain how to sketch the graph on an absolute value function and apply transformations at the form above...

    Whoever makes me understand gets 10wls
    Found this XD
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    Well rip I don’t this. But I do want to ask a question, when is the snake art contest going to be held.

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    Quote Originally Posted by Lopyhupis View Post
    Help,

    y=a*|1/b(x-h7)|+k

    I need help, someone explain how to sketch the graph on an absolute value function and apply transformations at the form above...

    Whoever makes me understand gets 10wls
    a is vertical stretch
    b is horizontal stretch
    h is left/right shift
    k is up/down shift

    first do vertical stretches and horizontal stretches
    then do left/right and up/down movements

    for graphing you can just run some numbers into for each x value to correspond to a y value or use desmos.
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    Master Sorcerer Yell0wTail's Avatar
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    You can't really graph the graph without any points/ not knowing the values of a,b, x, or y.

    (I think I'm not sure. I'm only in Algebra 2 lmao)

    If you have the numbers, I would glad to help
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    Master Sorcerer Lopyhupis's Avatar
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    Quote Originally Posted by Yell0wTail View Post
    You can't really graph the graph without any points/ not knowing the values of a,b, x, or y.

    (I think I'm not sure. I'm only in Algebra 2 lmao)

    If you have the numbers, I would glad to help
    Here It is, The Transamerica Pryamid is an office building in San Francisco. It stands 853 feet tall and is 145 feet wide at its base, Each unit represents one foot. Write an absolute value function whose graph is the V-Shaped outline of the sides of the building ignoring the “Shoulder”



    Here is another equation completely unrelated to any of the other problems on the HW but I have no idea how to do it, Name:  4D70B774-D9F7-4616-94E0-E0D32B2E816C.png
Views: 82
Size:  60.6 KB

    - - - Updated - - -

    Quote Originally Posted by SmartWaffle View Post
    a is vertical stretch
    b is horizontal stretch
    h is left/right shift
    k is up/down shift

    first do vertical stretches and horizontal stretches
    then do left/right and up/down movements

    for graphing you can just run some numbers into for each x value to correspond to a y value or use desmos.
    But, what if K is an Quadratic equations EG, g(x)=3/4|(x-5)|+5x^2 – 34x + 24

    Help me please lol I’m dying...
    Found this XD
    Quote Originally Posted by Seth View Post
    My bad, forgot Growtopia is a mobile game as well.
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  6. #6
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    Default My take on the question

    Hi, welcome to the world of absolute functions. To start of this explanation you must understand that absolute value functions are piecewise functions, meaning it is not continuous and is made out of different functions at different x coordinates.

    Let's start of with the easiest example:
    How does the graph of y = |x| look like? Well as you can see here: https://www.desmos.com/calculator/9egjbt3cug , the graph of y = |x| is comprised of two different functions.
    At x < 0, the function is y = -x, whereas at
    x ≥ 0, the function is y = x

    You can see that this is so here:
    https://www.desmos.com/calculator/ysnigdkjso

    Why is this so? This is because when x is positive {x≥0}, |x| will obviously be the same as x, however when x is negative {x<0}, |x| will equal -x, why? lets say we want to find the absolute value of a negative number, let's say -1... the |-1| is equal to -(-1) which of course equals to 1. The negative sign cancels the negativity of the x.

    Now lets try a slightly harder graph: y = |x-2| + 3
    in this case, when the inside of the absolute value is positive: x-2 ≥ 0 -> x ≥ 2, the graph will be y = x-2+3 -> y = x+1
    However, when the inside of the absolute value is negative: x-2 < 0 -> x<2, the (x-2) must be multiplied by -1 to stay positive, hence: y = -(x-2)+3 -> y=-x+5

    You can see that this is indeed the case here: https://www.desmos.com/calculator/5jgy9m3dlw (you can press the squiggly graph circle icon to the left of the functions to hide them, you can see that the graphs y = x+1 {x≥2} and y=-x+5 {x<2} are completely overlapping the graph y=|x-2|+3)

    Now that we have learnt the basics, we can now go to the question, how does the graph of y = a |1/b (x-h)| + k look like?
    Well again we make it to a piecewise function:
    1. When 1/b (x-h)≥0, |1/b (x-h)| = 1/b (x-h) -> When x≥h, |1/b (x-h)| = 1/b (x-h)
    2. When 1/b (x-h)<0, |1/b (x-h)| = -1/b (x-h) -> When x<h, |1/b (x-h)| = -1/b (x-h)


    hence:
    y=a/b (x-h)+k {x≥h} and y=-a/b (x-h)+k {x<h}

    a/b will control the gradient of the function because when expanded: y = (a/b)x + (k - ha/b), it follows the linear graph general formula y=mx+b where m is the gradient. If a increases, the graph will slant more and more (a is directly proportional to m), however if b increases, the function will get more flat (b is inversely proportional to m). You can see that this is the case if I equate a/b = m, a increases m as it increases, but as b increases, it divides a more, making the m decrease.

    h will control the x displacement of the graph. Why? this is because the (x-h). You can say that h minuses the x axis, making the x axis move to the left, hence as h increases, it's shifting the graph to the right, however as h decreases it shifts the x axis back to the right making the graph shifted to the left.

    k will control the y displacement, this is because if you put the k to the other side, you will get (y-k). With the same idea as before, minusing y by k will shift the y axis down, hence as k increases, the graph will be shifted up, however as k decreases, the graph will be shifted down because the y axis is getting more shifted back upwards.

    To understand this better I have provided sliders for you to play with the a, b, h, and k variables and see how changing them affects the graph:
    https://www.desmos.com/calculator/bp7wsiepc8

    Hope this is a clear explanation, you can ask again below if you have any questions
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    Quote Originally Posted by Lopyhupis View Post
    Here It is, The Transamerica Pryamid is an office building in San Francisco. It stands 853 feet tall and is 145 feet wide at its base, Each unit represents one foot. Write an absolute value function whose graph is the V-Shaped outline of the sides of the building ignoring the “Shoulder”



    Here is another equation completely unrelated to any of the other problems on the HW but I have no idea how to do it, Name:  4D70B774-D9F7-4616-94E0-E0D32B2E816C.png
Views: 82
Size:  60.6 KB

    - - - Updated - - -



    But, what if K is an Quadratic equations EG, g(x)=3/4|(x-5)|+5x^2 – 34x + 24

    Help me please lol I’m dying...
    That picture is an argand diagram, which is like a cartesian plane (x-y plane) however an argand diagram replaces x as the real axis and y as the imaginary axis (Re-Im plane). If you don't know what an imaginary number (i) is, i is basically the square root of negative one (√-1). It is represented as i to act more as a number than a square root because some math logic we have such as √a √b = √ab doesn't apply for negative numbers, i.e. √-1 √-1 = -1 and NOT √(-1)(-1) = 1

    Anyway, if you want to plot a point on the real axis (x axis), just write the number normally. However if you want to plot a point on the imaginary axis (y axis) you need to multiply it by i (it's like a unit for imaginary numbers, for real numbers the unit is 1, that's why there's no change). So overall you can define the point (a, b) on your x-y plane as a+bi in a Re-Im plane (argand diagram).

    But, why plus? Why should we add a with bi? well this is because a+bi acts like a vector, it is actually the vector from the origin O (0, 0) to the point (a, b). If you don't know what vectors are, it is basically something with BOTH a direction (it is pointing somewhere) and a magnitude (it has some value), some examples of vectors are force, velocity, acceleration, momentum, etc. (For example force is a vector because when you apply force on an object, let's say when you're pushing a table, you're applying force in a forwards DIRECTION, at a certain MAGNITUDE (For example, 15 Newtons, where Newton is the unit of force) to the table. Vectors are usually represented like an arrow, so for example "->" can mean 1 Newton to the right, or "<--" can mean 2 Newton to the left, however what if we want to represent vectors more generally instead of using words like left, right, up, down, etc. we can use the x, y, and z axis, which are represented by unit vectors (vectors in a certain direction with magnitude of 1) i, j, and k respectively. So if you were applying 5 N force to the right, you can say that it's (5i + 0j + 0k) N, or if you applied a 2 N force forwards, to the left, and to the north, it can be said as (2i - 2j + 2k) N. However, what if we want to find the magnitude of this (2i - 2j + 2k) N force? Well we can picture an arrow from the origin O (0, 0, 0) directed at 2 to the positive x axis, 2 to the negative Y axis, and 2 to the positive Z axis (2, -2, 2) and with pythagorean theorem we can see that the magnitude of this arrow is √(2²+2²+2²) which is 2√3, hence the magnitude of the force is 2√3 Newtons.

    Now, back to the Re-Im plane, a+bi can be thought of as an arrow from the origin (0, 0) to (a, b). Using pythagorean theorem the distance of this arrow is √(a²+b²), which is called the modulus. Now the higher the modulus, that means the larger the magnitude of the arrow/vector, hence the answer is Z1, because it is the furthest from the origin.







    Now for the second confusion, what if k is quadratic. Well first off, k should be a constant, not a function, however if you wanted to change k to that quadratic equation, again do the piecewise function separation I explained at my previous post, and you'll be able to graph it easily.
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    Master Sorcerer Lopyhupis's Avatar
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    Quote Originally Posted by Kelfran Gt View Post
    That picture is an argand diagram, which is like a cartesian plane (x-y plane) however an argand diagram replaces x as the real axis and y as the imaginary axis (Re-Im plane). If you don't know what an imaginary number (i) is, i is basically the square root of negative one (√-1). It is represented as i to act more as a number than a square root because some math logic we have such as √a √b = √ab doesn't apply for negative numbers, i.e. √-1 √-1 = -1 and NOT √(-1)(-1) = 1

    Anyway, if you want to plot a point on the real axis (x axis), just write the number normally. However if you want to plot a point on the imaginary axis (y axis) you need to multiply it by i (it's like a unit for imaginary numbers, for real numbers the unit is 1, that's why there's no change). So overall you can define the point (a, b) on your x-y plane as a+bi in a Re-Im plane (argand diagram).

    But, why plus? Why should we add a with bi? well this is because a+bi acts like a vector, it is actually the vector from the origin O (0, 0) to the point (a, b). If you don't know what vectors are, it is basically something with BOTH a direction (it is pointing somewhere) and a magnitude (it has some value), some examples of vectors are force, velocity, acceleration, momentum, etc. (For example force is a vector because when you apply force on an object, let's say when you're pushing a table, you're applying force in a forwards DIRECTION, at a certain MAGNITUDE (For example, 15 Newtons, where Newton is the unit of force) to the table. Vectors are usually represented like an arrow, so for example "->" can mean 1 Newton to the right, or "<--" can mean 2 Newton to the left, however what if we want to represent vectors more generally instead of using words like left, right, up, down, etc. we can use the x, y, and z axis, which are represented by unit vectors (vectors in a certain direction with magnitude of 1) i, j, and k respectively. So if you were applying 5 N force to the right, you can say that it's (5i + 0j + 0k) N, or if you applied a 2 N force forwards, to the left, and to the north, it can be said as (2i - 2j + 2k) N. However, what if we want to find the magnitude of this (2i - 2j + 2k) N force? Well we can picture an arrow from the origin O (0, 0, 0) directed at 2 to the positive x axis, 2 to the negative Y axis, and 2 to the positive Z axis (2, -2, 2) and with pythagorean theorem we can see that the magnitude of this arrow is √(2²+2²+2²) which is 2√3, hence the magnitude of the force is 2√3 Newtons.

    Now, back to the Re-Im plane, a+bi can be thought of as an arrow from the origin (0, 0) to (a, b). Using pythagorean theorem the distance of this arrow is √(a²+b²), which is called the modulus. Now the higher the modulus, that means the larger the magnitude of the arrow/vector, hence the answer is Z1, because it is the furthest from the origin.







    Now for the second confusion, what if k is quadratic. Well first off, k should be a constant, not a function, however if you wanted to change k to that quadratic equation, again do the piecewise function separation I explained at my previous post, and you'll be able to graph it easily.
    Aight I get it now TYSM! World name?
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    Quote Originally Posted by Lopyhupis View Post
    Aight I get it now TYSM! World name?
    World Name: KELFRAN
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    Master Sorcerer Lopyhupis's Avatar
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    Quote Originally Posted by Kelfran Gt View Post
    World Name: KELFRAN
    Aight I deposited it

    Name:  70364695-43F7-4E62-949A-D1F49147C4C7.jpeg
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    Last edited by Lopyhupis; 09-25-2019 at 03:43 PM.
    Found this XD
    Quote Originally Posted by Seth View Post
    My bad, forgot Growtopia is a mobile game as well.
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    https://www.growtopiagame.com/forums...t-1Dl-(Snakes)

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