# Finding the Parameters for the Curve y=asin(x-b)+c

• Natasha1
In summary, the member has been informed to separate different questions into different threads in the future. The conversation revolved around finding the coordinates of the minimum point of a curve with different equations, as well as finding the values of variables a, b, and c in a graph. The correct values for both parts (a) and (b) were provided, with explanations for why they are correct. The member was also reminded to post each part of the problem in separate threads.

#### Natasha1

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

1. Homework Statement

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

#### Attachments

• Screenshot 2019-03-08 at 19.19.51.png
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Natasha1 said:

## Homework Statement

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

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):
Graph for part (b):

#### Attachments

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Natasha1
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) ?

SammyS said:
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 239951Graph for part (b):
View attachment 239952
SammyS said:
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

Natasha1 said:
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 ?

x = 3

x = 3

Natasha1 said:
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) .

Natasha1
Thanks SammyS but what about the rest of my work?

Natasha1 said:
Thanks SammyS but what about the rest of my work?

(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) .

(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)

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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## What is the minimum point of a curve?

The minimum point of a curve is the point where the curve reaches its lowest value. It is also known as the local minimum or the global minimum, depending on the context.

## How is the minimum point of a curve calculated?

The minimum point of a curve is typically calculated by finding the derivative of the curve and setting it equal to zero. This will give the x-coordinate of the minimum point, which can then be plugged into the original equation to find the y-coordinate.

## What does the minimum point of a curve represent?

The minimum point of a curve represents the lowest value that the curve can attain. It is often used in optimization problems to find the most efficient solution.

## Can a curve have more than one minimum point?

Yes, a curve can have multiple minimum points. These points can be either local or global minimums, depending on the shape of the curve.

## How is the minimum point of a curve used in real life?

The concept of the minimum point of a curve is used in many real-life applications, such as finding the lowest cost or the most efficient route in transportation, determining the optimal production level in manufacturing, or identifying the most profitable investment strategy in finance.