Transformations


Learning Objectives

 

Problem 1 of 1

 

Given the graph of y = ƒ(x) at right, sketch a clearly labeled graph of each of the following:

 

(a) y = 2ƒ(x)

 

(b) y = ƒ(2x)

 

(c) y = -ƒ(x)

 

(d) y = ƒ(-x)

 

Solution

- 2 stretches the graph veritically,

which means...

 

- multiply the y-coordinates of f(x) by 2

 

(2*2 = 4)

 

therefore the new y-coordinate is 4.

Since the y-cooridinates are only changed,

the x coordinates remain the same

 

 

The graph will be compressed by a

factor of 1/2,

 

because we use the reciprocal of

b (b is "2" in the equation)

 

Also, this is a compression, which means

the x coordinates will only be affected.

- so multiply the two x-coordinates of the graph

by 1/2

 

3*1/2 = 3/2

 

-3*1/2 = -3/2

 

Therefore the new x-coordinates

are 3/2 and -3/2

the y-coordinate remains the same

 

 

 

The negative in front of the equation,

creates a reflection on the x- axis.

 

Therefore, the y-coordinates get multiplied by (-1)

 

2*(-1) = -2

 

The new y-coordinate is -2

The x-coordinates are unaffected and remain the same

 

 

The negative in front of the "x", causes all x coordinates

to be multiplied. Negative x-coordinates creates a

reflection across the y-axis

 

Therefore the x-coordinates get multiplied by (-1)

 

3*(-1) = -3

 

-3*(-1) = 3

therefore there is no visual change, because the shape of the

function is the same. The reflection still occurs, but is unnoticeable

the reflection is only noticeable in the equation.

 

The y-coordinates are unaffected by this transformation.

 

 

 

Problem 2 of 3

 

Given the graph of y = ƒ(x) at left, sketch a clearly labeled graph of each of the following:

 

(a) y = ƒ(x + 3)

 

(b) y = ƒ(x) - 2

 

(c)

 

Solution

(a)

 

· The general equation for a tranformation is expressed as y = ƒ(x - a)+ b.

 

· In this transformation of y = ƒ(x + 3). The role of parameter a is being affected.

· a affects the translation of the graph horizontally (x-axis). Watch the negative sign, because in the general equation there is a negative.

· In this case it is translated left 3 units because it is really y = ƒ(x - (- 3)) from following the general equation.

· Therefore, add -3 to all the x-coordinates of the graph to shift it left.

 

If a < 0 the graph shifts left a units.

If a > 0 the graph shifts right a units.

 

(b)

 

· In this transformation of y = ƒ(x) - 2. The role of parameter b is being affected.

· b affects the translation of the graph vertically (y-axis). There is no need to watch out for the signs for this one.

· In this case it is translated downwards 2 units.

· Therefore, add -2 to all of the y-coordinates of the graph to shift it down.

 

If b < 0 the graph shifts down b units.

If b > 0 the graph shifts up b units.

 

(c)

 

· To answer this kind of problem we must remember that we are working on the outputs of y-coordinates.

· First identify the vertical asymptote which always lay on the y-axis and the roots (whenever y = 0).

· Then identify the invariant points, these are the points that do not move which is when y=1 and y=-1.

· Then identify if f is increasing negatively, then 1/f(x) is decreasing negatively then vice versa.

 

If 1/f(x)>1, this means that f is decreasing If 1/f(x) is increasing and vice versa

 

Problem 3 of 3

 

Given the graph of y = g(x) below:

 

(a) Sketch the graph of y = -g(x) + 3

 

(b) Sketch the graph of y = g-1(x)

 

(c) Sketch the graph of

 

Solution