# Minimum point of a curve

<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
89.6 KB · Views: 588
Last edited by a moderator:

SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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

• Natasha1
SammyS
Staff Emeritus
Homework Helper
Gold Member
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) ?

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
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

SammyS
Staff Emeritus
Homework Helper
Gold Member
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?

SammyS
Staff Emeritus
Homework Helper
Gold Member
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)