Functions  Transformations.
Test Yourself 1  Solutions.
Remember when determining SHIFTS, we need to change a given equation into the form
When determining DILATIONS, we need to change a given equation into the form
Also remember that it is generally better to do the dilations before doing the shifts.
Shift only  1. When y = x^{2} is shifted 2 units left:
y = (x + 2)^{2} 
2. When y = x^{3} is shifted 3 units right:
y = (x  3)^{3} 
3. When y = 5^{x} is shifted 5 units down:
y + 5 = 5^{x} or y = 5^{x}  5 
4. When y = x  5 is shifted 2 units up:
y  2 = x  5 or y = x  3 

5. Shifting 3 units to the left
changes y = 5x^{2} + 2 to y = 5(x + 3)^{2} + 2 
6.
Shifting 0.5 units up changes
y = 2√x  2.6 to y  0.5 = 2√x  2.6? So y = 2√x  2.1. 

7. Shifting 6 units down changes
y = 5 to
y = 1. 
8. Shifting 2 units to the right changes x = 4 to x = 2. 

Dilation only  9. Dilating y = x^{2 }vertically by a factor of 2 gives y/2 = x^{2} or y = 2x^{2}.  10. Dilating y = 2^{x} vertically by a factor of 1/3 gives 3y = 2^{x} (after dividing y by 1/3). 
11. Dilating horizontally by a factor of 3 gives .  12. Dilating y = x^{2} horizontally by a factor of 0.5 gives  
13. The original parabola has x intercepts at x = 2 and x = 1 and a vertex at (0.5, 2.25). Dilating y = (x  1)(x + 2) horizontally by a factor of 3 (i.e. expanding or dilating it horizontally) but no shift requires dividing the x terms in BOTH bracketed terms by 3: So both the vertex and the x intercepts are dilated  i.e. multiplied by 3  so the x intercepts change to x = 6 and x = 3 (i.e. the original values are multiplied by 3) and the new vertex is at(1.5, 2.25). 
14. Dilating
y = (x  1)(x + 2) vertically by a factor of 3 gives
Hence the x intercepts remain at x = 2 and x = 1. The x value for the vertex remains at The parabola rises three times as fast as previously. 

Identify the shift  15. To track the change of y = 2x^{2} + 1 to y = 2x^{2}  3 we rewrite to isolate the original equation: y = (2x^{2} + 1)  4. ∴ y + 4 = (2x^{2} + 1) So there is a vertical shift of 4. All y values are reduced by 4. 
16. To track the change of y = 4x^{3} to y = 4 (x + 2)^{3}, no rewriting is necessary as it is in the relevant format. There is a horizontal shift of 2. For example the x intercept changes from

17. To change y = x^{2} + y^{2} to
y = (x + 1)^{2} + (y  2)^{2} there is a horizontal shift of 1 and a vertical shift of 2. 
18. To change y = sin x to y = sin (x  π) requires a horizontal shift of π units to the right. 

Identify the dilation  19. To transform y = 2x^{2} to y = 4x^{2}, rewrite as y = 4x^{2} = 2(2x^{2}).
Hence the effect is a vertical dilation of 2 which stretches all y values by a factor of 2 (i.e. multiplies all y values by 2). 
20. To transform
y = 3 cos x to y = cos x:
The effect of this transformation is to reduce all y values by one third. So the x intercepts remain the same but the maximum values reduce from 3 to 1. 
21. To transform y = 2^{x} to y = 2^{x+2}:
The effect of the vertical dilation is to change all y values by a dilation factor of 4 (i.e. to multiply all y values by 4  see especially x = 0). 
22. To transform
y = 2x + 4 to y = x + 2:
The effect of the vertical dilation is to halve all y values for each x value. Hence the point (2, 8) on y = 2x + 4 becomes (2, 4) on y = x + 2. 

Shift and dilation  23. To change x^{2} + y^{2} = 4 into (x  3)^{2} + (y + 4)^{2 }= 16:
The centre of the original circle is (0, 0). The dilation transformations change the radius from 2 to 4. The shift transformations move the centre from 
24. The transformation(s) to change x^{2} + y^{2} = 1 into require rewriting the equation as:
The circle turns into a horizontal ellipse.

25. To transform y = cos x into
y = 5cos(2x  π/3): So the transformations are:

26. To transform the equation
to :
Start with the vertical dilation: So the transformations are:
The peak of the semicircle is at (0, 1):


27. To transform the equation
to become : It is possible to start transforming the original equation by incorporating the vertical shift: Hence:
Note in the above graph:

28. Starting with the equation
what transformations change the equation to ? Transformations are:


Use the following diagram of y = f(x) = x^{2} to sketch new graphs in response to the information given in each of questions 2930.  
29.
Note: There is a horizontal shift of 3 to the left and a horizontal dilution (expansion) by a factor of 3 (so the curve flattens out). Hence the parabola flattens and the vertex moves to x = 3. There is a vertical shift of 2. Hence the vertex is now at (3, 2) rather than (0, 0). At x = 0, y = 1 + 2 = 3. 
30.
Note: there is a horizontal shift of 2 to the right (so the vertex is now at x = 2) and a horizontal dilution of 1/3 (hence the curve becomes narrower). There is a vertical shift of 1 upwards (so the vertex moves to y = 1) and a vertical dilution by a factor of 2 (hence the curve climbs more quickly). 

Use the following diagram of y = f(x) to sketch new graphs in response to the information given in each of questions 3134.  
Diagrams  31.
Note: The horizontal shift of 1 moves the previous point at x = 1 to the left and it is now on the y axis. The minimum points at about x = 0.5 and x = 5.75 have moved left to about x = 1.5 and to about There is no vertical change. 
32.
Note: The previous centre of the loops was at The horizontal shift of 2 moves the maximum point from about x = 2.5 to the left to x = 0.5. The next minimum at about 3.5 has therefore been shifted left to about 1.5. 
33.
Note: The previous centre of the loops was at about The vertical shift of 1 moves the centre between the peaks and troughs from y = 2 to y = 3. The vertical dilation of 2 changes one minimum value to y = 0 at x = 0.5 and one maximum value to y = 6 at x = 2.5 respectively (a difference of 6 which is twice the original amplitude). There is no horizontal shift or dilation. 
34.
Note: There is a horizontal dilation of ½ which compresses the pattern. Previously one of the lowest points occurred at x = ½ whereas now, the corresponding lowest point is at x = ¼. The distances between the two troughs was originally 6 units but is now 3 units. The distance above the y axis was originally ½ but it is now 1. The vertical dilation of a factor of 2 increases the amplitude of thepattern from 3 units to 6 units. 