Minimum point of a curve

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<Moderator's note: Member has been informed to separate different question into different threads in the future.>

1. Homework Statement

Please see picture attached...
The diagram shows the curve with equation y = f (x)
The coordinates of the minimum point of the curve are (–2, –1)

(a) Write down the coordinates of the minimum point of the curve with equation

(i) y=f(x–5)
(ii) y = 0.5 f ( x )

The graph of y=asin(x–b)+c (please see graph on picture)
b) Find the value of a, the value of b and the value of c.

The Attempt at a Solution


(a) Write down the coordinates of the minimum point of the curve with equation

(i) y=f(x–5) Does the -5 move the graph 5 units to the right? Why is this? So the coordinates are (-7, -1)
(ii) y = 0.5 f(x) Does the times 0.5 widen the curve and has no impact on the minimum point so would stay (-2, 1)

b) Find the value of a, the value of b and the value of c.
Is a the magnitude? so a = 1
Is b the phase shift so -b shifts the graph by b units to the right so b = 360
Is c the vertical shift which is c = -1

Please explain :). Thank you in advance...
 

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Answers and Replies

  • #2
SammyS
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Homework Statement


Please see picture attached...
The diagram shows the curve with equation y = f (x)
The coordinates of the minimum point of the curve are (–2, –1)

(a) Write down the coordinates of the minimum point of the curve with equation

(i) y=f(x–5)
(ii) y = 0.5 f ( x )

The graph of y=asin(x–b)+c (please see graph on picture)
b) Find the value of a, the value of b and the value of c.

The Attempt at a Solution


(a) Write down the coordinates of the minimum point of the curve with equation

(i) y=f(x–5) Does the -5 move the graph 5 units to the right? Why is this? So the coordinates are (-7, -1)
(ii) y = 0.5 f(x) Does the times 0.5 widen the curve and has no impact on the minimum point so would stay (-2, 1)

b) Find the value of a, the value of b and the value of c.
Is a the magnitude? so a = 1
Is b the phase shift so -b shifts the graph by b units to the right so b = 360
Is c the vertical shift which is c = -1

Please explain :). Thank you in advance...
Even though these are shown as parts (a) and (b) on the image, the parts have almost nothing in common and it would make sense to post each part in a separate thread.
Graph for part (a):
upload_2019-3-8_14-13-38.png



Graph for part (b):
upload_2019-3-8_14-15-33.png
 

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  • #3
SammyS
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Hint for Part (a):
You should know the value of ƒ(−2). Right?

So, what value must x have in order for ƒ(x − 5) to give the value of ƒ(−2) ?
 
  • #4
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Even though these are shown as parts (a) and (b) on the image, the parts have almost nothing in common and it would make sense to post each part in a separate thread.
Graph for part (a):
View attachment 239951


Graph for part (b):
View attachment 239952
Hint for Part (a):
You should know the value of ƒ(−2). Right?

So, what value must x have in order for ƒ(x − 5) to give the value of ƒ(−2) ?

ƒ(−2) = -1
So, what value must x have in order for ƒ(x − 5) to give the value of ƒ(−2) ? I don't know ƒ(x − 5) = -1
 
  • #5
SammyS
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ƒ(−2) = -1
So, what value must x have in order for ƒ(x − 5) to give the value of ƒ(−2) ? I don't know ƒ(x − 5) = -1
Then, what value must x have so that (x − 5) = −2 ?
 
  • #8
SammyS
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x = 3
Right.
So, if you are graphing ƒ(x−5), then when x = 3, y = ƒ(3 − 5) = ƒ(−2) = −1 , does it not?

Thus, the graph of y = ƒ(x−5) is just like the graph of y = ƒ(x), except that
the graph, y = ƒ(x−5) is shifted 5 units to the right as compared to the graph of y = ƒ(x) .
 
  • #9
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Thanks SammyS but what about the rest of my work?
 
  • #10
SammyS
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Thanks SammyS but what about the rest of my work?
Most of your answers are incorrect.

(a) : (i)
It is true that the graph, y = ƒ(x − 5), is obtained from the graph, y = ƒ(x), by shifting the latter by 5 units to the right, which is what you said, but then your answer of (−7, −1) for the coordinates of the vertex (a minimum) is incorrect. You need to shift the graph, not the coordinate system. Also, I gave you some idea of how to think of this as a shift.

(a) : (ii)
While it is true that multiplying ƒ(x) by 0.5 gives a graph that appears to be widened (at least for this function), what multiplying by 0.5 actually does is to "shrink" the graph vertically. This is because for any particular x value, the y value for y = 0.5ƒ(x) is one half of the y value for y = ƒ(x) .

Part (b) answers are incorrect.
 
  • #11
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Are the answers correct please?

(i) y=f(x–5) Does the -5 move the graph 5 units to the right? Why is this? So the coordinates are (3, -1)
(ii) y = 0.5 f(x) Does the times 0.5 widen the curve and has no impact on the minimum point so would stay (-2, -1)
 
  • #12
493
8
Are the answers correct please?

b) Find the value of a, the value of b and the value of c.
Is a the magnitude? so a = 3
Is b the phase shift so -b shifts the graph by b units to the right so b = 60
Is c the vertical shift which is c = 0 (as there is no vertical shift)
 
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