Read the paper, Conceptions of function translation: Obstacles, intuitions, and rerouting, by Zazkis, Liljedahl, and Gadowsky.

1. Explain in your own words the input-output explanation for why the graph of y=(x+3)^2 is a horizontal shift three units to the left. Note that this explanation does not appear in the paper.

2. Write a half page, single-spaced reflection on the article. What you choose to reflect on from the article is up to you, but I suggest that you reflect on aspects of the reading that resonated with you, that you found most interesting or most illuminating for your own mathematical understanding.

3.   Interview a high school student, college student, high school teacher, or college instructor (basically interview one person that you would expect to know a little something about functions). Conduct your interview over zoom. Record the interview to the cloud and change you settings to get an automatically generated transcript. Use the same interview protocol as in the paper. In the paper pay close attention to the dialogue between the interviewer and the interviewee for the kind of probing questions that the interview asks. Strive to do the same in your interview.

Interview protocol

Start by inviting the participant to sketch by hand on the same coordinate system graphs of y=x^2 and y=(x-3)^2 and ask them to explain how they generated their sketch.

• Next, invite participant to compare their sketch to that of computer generated sketches (use Desmos or another web based graphing program).
• After they compare sketches, ask the participant to comment on the location of the graph y = (x – 3)^2 relative to the graph of y = x^2, relating, where necessary, to the discrepancy between participants sketch and the computer generated sketch, or to the discrepancy between the intuitive expectation and the “known” result.
• If the issue did not come up naturally, ask the participant to discuss the graph of y = x^2? 3 and compare it to the graph of y = (x – 3)^2.

Write a one page max, single-spaced account of the mathematical reasoning of the person you interviewed. Use exact quotes (like they did in the paper). After summarizing their reasoning do your best to “diagnose” their reasoning and compare their response to those in the paper.

wA shift of the graph up, down, left, or right, without changing the shape, size, or dimensions of the graph, is called a translation. (I heard this definition from Nina in my group 4) which remind me how we memorized before the rule of this events without going deeper. This motivate me rethinking about something related to this events, it is “stretch or shrink the graph” Which usually changes the shape of graph.

Adding to the input increases the function in the y direction, adding to the input decreases the function in the x direction. This is because the function must compensate for the added input. If the function outputs “7” when “3” is input, and we input x + 2, the function will output “7” when x = 1.

Examples: If f (x) = x2 + 2x, what is the equation if the graph is shifted:

4 units up f1(x) = f (x) + 4 = x2 + 2x + 4

4 units down f2(x) = f (x) – 4 = x2 + 2x – 4

4 units left f3(x) = f (x + 4) = (x + 4)2 +2(x + 4) = x2 +8x + 16 + 2x + 8 = x2 + 10x + 24

4 units right f4(x) = f (x – 4) = (x – 4)2 +2(x – 4) = x2 -8x + 16 + 2x – 8 = x2 – 6x + 8

A shift of the graph up, down, left, or right, without changing the shape, size, or dimensions of the graph, is called a translation.

To stretch or shrink the graph in the x direction, divide or multiply the input by a constant. As in translating, when we change the input, the function changes to compensate. Thus, dividing the input by a constant stretches the function in the x direction, and multiplying the input by a constant shrinks the function in the x direction.

We can understand the difference between altering inputs and altering outputs by observing the following:
If g(x) = 2f (x): For any given input, the output of g is two times the output of f, so the graph is stretched vertically by a factor of 2.
If g(x) = f (2x): For any given output, the input of g is one-two the input of f, so the graph is shrunk horizontally a factor 2.

EXAMPLE

y=f(x)

1.y=2f(x)

vertical stretch: – y-values are doubled; points get farther away from x-axis.

1. y=f(x)/2

vertical shrink: y-values are halved; points get closer to x-axis.

1. y=f(2x)

horizontal shrink: x-values are halved; points get closer to y-axis

y=f(x/2): – horizontal stretch; x-values are doubled; points get farther away from y-axise explain like that: 4 attachmentsSlide 1 of 4

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### UNFORMATTED ATTACHMENT PREVIEW

Question 1: When solving a differential equation, I will get an expression of x in term of variable t , which can further be denoted by x  t  . Question 2: (a) Using separation of variable for the given differential equation, we obtain: dy 4 t y dt dy    t 4 dt y  ln  y    ye t5 C 5 t5 C 5 t5 5  y  ke , k  eC (b) Using separation of variable for the given differential equation, we obtain: dy  2 y 1 dt dy   dt 2 y 1  1  ln  2 y  1  t  C 2  ln  2 y  1  2t  C   2 y  1  e 2t C  1 1  y  ke 2t  , k  eC  2 2 (c) Using separation of variable for the given differential equation, we obtain: dy t3 y dt dy   1/3   tdt y 3 y 2/3 t 2  C 2 2 2 t  y 2/3   C  3 variable for the given differential equation, we obtain:   t2   y    C  3  3/2 Question 3: (a) Using separation of variable for the given differential equation, we obtain: dy t  dt y   ydy    tdt y2 t2   C 2 2 2 2  y  t  C   Plugging y  0   4 into the above equation gives y 2  t 2  C   42  02  C   C   16 Hence, it follows that y 2  t 2  16 . Since y  0   0 , we can assume y  t   0 for all t , and the equation can be solved explicitly as follow: y 2  t 2  16  y  t 2  16 (b) Using separation of variable for the given differential equation, we obtain: dy  3 y dt   y 1/3 dy    dt  3 y 2/3  t  C 2 Plugging y  0   27 into the above equation give 3 y 2/3  t  C 2 3  27 2/3  0  C s 2 27 C  2 We have 3 y 2/3 27  t  , which can be solve explicitly as follow: 2 2 3 y 2/3 27  t  2 2 2t  y 2/3    81 3  2t   y     81  3  3/2 which is the required particular solution. Question 4: dy y   y 1   . It implies that y  6 is the solution dt  6 dy d because both sides of the differential equation are equal to 0 . In fact,   6   0 and dt dt y   6 y 1    6 1    6  0   0 . On the other hand, y  8 is not the solution because we still  6  6 dy d y 8   8 have   8   0 but y 1    8 1     which is not 0 . dt dt 3  6  6 The required differential equation is Question 5: The given graph represents the solution of the given differential equation P  t   C  e0.2t , where C is the constant. dP  0.2 P , which is dt The given differential equation is solved by: dP  0.2 P dt dP   0.2  dt P  ln P  0.2t  C  P  e0.2t C  P  C e0.2t Question 6: dy  y  t is read with meaning, and it indicates that the rate of dt change of y is equal to the difference between y and t . (a) The differential equation (b) If the functions (i)-(iv) are indeed the solution to the given differential equation then the vector in the slope field of dy  y t , dt dy  y  t must be tangential to the curve of the functions dt (i)-(iv) at any time t . t (c) We have that (iii) y  t   e  t  1 is the only solution to dy  y  t because dt dy d t   e  t  1 dt dt dy   et  1 dt dy    et  t  1  t dt dy   y t dt Other than that, the functions (i), (ii), and (iv) are not the solutions to the given DE dy  y t . dt Question 7: (a) The relevant terms in the glossary are:      initial value problem (IVP) differential equation separation of variables slope field solution to a differential equation (exact, particular, general, explicit, implicit) (b) There are no other terms that may be relevant for this unit but were not included. Group 4 translations – Google IX BSU9ZU0az1ZjY5220VW2mphebohleditsidealdgdce7b3e505 mnge Tools Add-ons Help Last edit was minutes ago CONOC Background Layout Theme Transition S3 SERRALHERIA Use an input-output meaning of function to explain why the graph of y, = (x+3)2 is a 3 unit shift left of the graph of y =x². Suggestion: consider when the output of y, is the same as the output of y,. We don’t have to think of the shift as just “the opposite of what it looks like”. We can reason that to maintain equality with the left side of the equation (the output, y) the x must get smaller by 3 to maintain the same value from before. We know 16=(4)^2 [y=x^2}, now to keep the 16, we have 16=(1+3)^2, [y=(x+3)^2], see how the x went from 4 to 1? It got smaller by 3, a shift left of 3. The quantity being squared has to stay 4. So when x changes to x+3, the x goes from 4 to 1. ker notes Ae O 39 MO Pause/Stop Recording Participants Chat Share Screen Reactions DELL F FO FB F9 F10 3 7 8 R D F G нЈ K 7. В мм Verify that L(t) = 1 – 2-0.5(t + c) is the general solution to the rate of change equation dL/dt = 0.5(1-L). Next, explain why the graph of L = 1 – e-0.5(t + 3) is a 3 unit horizontal shift left of the graph of L, = 1 – 0.5€. The solution is verified as follow: dL/dt = d/dt(1 – e-0.5(t + c)) = 0.5e-0.5(t + c) = 0.5 (1-(1 – e-0.5(t + c))) = 0.5(1-L), as desired result. We denote f(t)= 1 -e-0.5t. It is known that

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