Mechanics Transformations

Question

A) What are the domain and range of f(x)? [Note the arrow on the left side and the closed endpoint on the right side]

B) Label at least 5 key points on the graph. It helps to write them in a table. 

C) The original graph is transformed - it is shifted 2 units to the right and shifted 2 units up. Label where your 5 points have moved to, and draw a graph of the transformed function. What is the new formula for the transformed function in terms of f(x)? What are the new domain and range of the transformed graph? 

D) The original graph is transformed - there is a horizontal stretch by a factor of 2, and a vertical stretch by a factor of 2. Label where your 5 points have moved to, and draw a graph of the transformed function. What is the new formula for the transformed function in terms of f(x)? What are the new domain and range of the transformed graph? 

E) The original graph is transformed - there is a horizontal reflection over the y-axis and also a vertical reflection over the x-axis. Label where your 5 points have moved to, and draw a graph of the transformed function. What is the new formula for the transformed function in terms of f(x)? What are the new domain and range of the transformed graph?

Answer 1

Hi Ximena,

This is a bit of a long one.

So for part A of the question:

The domain is any value that can be entered into the function (represented by the x value) and result in a solution (represented by the y value). Looking at the graph, we see that the highest x value that is defined for the function is at x = 6 and the lowest x value is -∞ (The arrow on the left side of the function means that the function continues forever in that direction).

The range is the highest and lowest values that the function can take given any x value that lies in the domain. Looking at the y axis, we can see that the lowest value y takes on is -2 and the highest is ∞ (represented by the arrow).

Therefore, the answer to A is

Domain = (-∞,6]

Range = [-2,∞)

It is important to note that when using interval notation, a parenthesis means that the number on that side is not included in the interval and a bracket means that it is. We can see that 6 and -2 are points on the graph, but while you can get closer and closer, -∞ and ∞ are numbers that can never be reached and therefore, are not included in the interval.

For part B, we can consider what points would be significant on this function.

Some important properties to point out could include the y-intercept, x-intercepts, local maximums, local minimums, absolute maximum, absolute minimum.

1. The y-intercept is the point where the function touches the y axis, which happens at P(0,2)
2. An x-intercept is a point where the function touches the x-axis (where y equals 0), which happens at P(-2,0), P(2,0), and P(6,0)
3. The local maximum is the highest point in a given interval. The local maximum between the interval [-2,2] is at P(0,2).
4. The local minimum is the lowest point in a given interval. The local minimum between the interval [2,6] is at P(4,-2).
5. The absolute maximum is the largest value (the largest y value) the function can take on. The absolute maximum is at P(-∞,∞). In this case, this is not considered a point, and therefore would not appropriately answer the question.
6. The absolute minimum is the smallest value (the smallest y value) the function can take on. The absolute minimum is at P(4,-2).

For part C, you just have to add the x and y values to any of the previous points mentioned.

When translating the function left or right, we are adding or subtracting a value to or from the x value of the function.

For example,

If you have the function f(x) = (x-3)² + 6

and move it 2 units to the right, you subtract 2 from x in the function to get:

(x-5)² + 6

This might seem counter-intuitive that a negative value would move the function in the positive direction, but it makes sense when you consider that the x value would need to be 2 greater than before in order to attain the same y values.

To move a function along the y axis, you simply take whatever y value the function would have taken on prior to translating it, and add or subtract a number from that value to move that number of units up or down.

If we move the previous expression 2 units up, we add two

[(x-5)² + 6] + 2

= (x-5)² + 8

Therefore, to move the function to the right 2 units and up 2 units, we would write the new function as,

f(x-2) + 2

Now, let's apply this to the points we mentioned. When we move each point by (+2,+2)

1. The y-intercept P(0,2) moves to P(2,4)
2. The x-intercepts become
3. P(-2,0) -> P(0,2)
4. P(2,0) -> P(4,2)
5. P(6,0) -> P(8,2)
6. The local maximum between the interval [-2,2] moves from P(0,2) to P(2,4) and the new interval is [0,4]
7. The local minimum between the interval [2,6] moves from P(4,-2) to P(6,0) and the new interval is [4,8]
8. The absolute minimum moves from P(4,-2) to P(6,0)

To get the new domain and range for the new function f(x-2)+2,

add 2 to the lower and upper limits of the domain and range f(x)

Domain = (-∞ + 2,6 + 2]

= (-∞,8]

Range = [-2 + 2,∞ + 2)

= [0,∞)

Negative or positive infinity does not change when adding, subtracting, multiplying, or diving any number to it. The exceptions are if you are dividing or subtracting by ∞ or multiplying by 0.

For part D, when you stretch a function along the x, you divide the x value by the stretch factor. When you stretch a function along the y-axis, you multiply the entire function by the stretch factor. If we stretch our new function by a factor of 2 in the x-direction and 2 in the y-direction,

after accounting for the horizontal stretch the function becomes: f((x-2)/2)² + 2

to add the vertical stretch, the function becomes:

2[f((x-2)/2)² + 2]

To get a 2x horizontal stretch, we divide the x value by 2, including the previous translation value. This may seem counter-intuitive, as well, but if you are dividing x by two it will take 2 times the x value to attain the same function value as before. This makes the axis "longer".

To stretch in the verticle direction by a factor of 2, you just have to multiply the entire function by 2 which will double the y value for each given x value.

To get the new domain and range we multiply the x and y values by their respective stretch factor

1. Domain = (-∞*2,8*2] = (-∞,16]
2. Range = [0*2,∞*2) = [0,∞)

For part E,

When reflecting over the y-axis, the new x value takes on the opposite sign. For example, a point in quadrant 2 has a positive y value and a negative x value. If that point is reflected over the y-axis, it now sits in the 1st quadrant with a positive y value and a positive x value. The negative x became a positive x.

When reflecting over the x-axis, the new y value takes on the opposite sign. For example, if you have a point in quadrant 3, both the x and y values will be negative. After a reflection over the x-axis, the point will now sit in the 2nd quadrant where the x value is negative and the y value is positive. The negative y became a positive y.

The question asks us to reflect over both the x and y-axis, which means that both the x and y values will take on the opposite sign after being reflected.

To change the sign of the y value, we multiply f(x) by (-1)

To change the sign of the x value, we multiply x by (-1) to get f((-1)x)

2[f((x-2)/2)² + 2]

becomes

-2[f((-x-2)/2)² + 2]

The new domain after reflecting:

1. Domain = (-∞(-1),16(-1)] = [-16,∞)
2. Range = [0(-1),∞(-1)) = (-∞,0]

note that the lower and upper bounds of each interval swapped places

Each point after reflecting

1. The original y-intercept now at P(2,4) becomes P(-2,-4)
2. The original x-intercepts become
3. P(0,2) -> P(0,-2)
4. P(4,2) -> P(-4,-2)
5. P(8,2) -> P(-8,-2)
6. For the local maximum P(2,4) in the interval [0,4], the new interval becomes [0,-4], and the new local maximum becomes P(-2,-4)
7. For the local minimum P(6,0) in the interval [4,8], the new interval becomes [-4,-8], and the new local minimum becomes P(-6,0)
8. The absolute minimum moves from P(6,0) to P(-6,0)

I hope that helped!

Answer 2

A) The Domain of a function is measured along the x-axis. Here the greatest value on the x-axis is marked as 6 with a filled-in circle indicating it is included in the function, and the arrow at the end of the graph indicates it goes on infinitely in the negative direction. An inequality notation of this function would be x≤6. You could also use an interval notation which would be (-∞, 6]. The Range of a function is measured on the y-axis. Here the lowest value on the y-axis is marked as -2, and the arrow at the end of the graph indicates it goes on infinitely in the positive direction. An inequality notation of this function would be y≥-2. You could also use an interval notation which would be [-2, ∞).

In both cases, the bracket represents the 'or equal to' of the inequality notation, otherwise, it would have been another parenthesis.

B) In this case there are some clear points on the graph. A point is always written (x, y). For example, the point marked on the graph is at (6, 0). You can apply this to find other points on the graph as well.

C) Shifting the graph up and to the right is moving it in the positive direction of both axes, meaning you can simply add two to all previous values. It may help to plot out the new points so you can see the new graph visually. For example, our new marked point would be at (8, 2), adding 2 to both the 6 and 0 of its previous location. This would shift the domain so that the greatest value was 8, and the range so that the lowest value is 0. You can use the previous notations, just replacing the previous numbers with the new numbers.

D) First we can work on the formula in terms of f(x). In order to stretch a graph along the y-axis, you would write the function as nf(x), in this case, 2f(x), where n represents the factor by which you are stretching. To stretch along the x-axis, you would write the function as f(x/n), in this case, f(x/2). We can see the difference when we try it out. If g(x)=2f(x), the output of g is double that of f, and so the graph is stretched vertically by a factor of 2. If g(x)=f(2x), the output of g would be half that of f, and so the graph is shrunk horizontally by a factor of 2. Our new formula in terms of f(x) would be 2f(x/2) since the graph is being stretched in both directions.

Next, we can work on the points. While stretching a graph, I would recommend doing each stretch one at a time and marking out the new points each time. It will be easier to visually understand the stretch this way. For this, I will start with our point (6, 0). For this, we simply multiply each value by 2, which would bring our new point to (12, 0). We can also use our point (0, 2), which would become (0,4). Both of these points were on axes, so they only moved in one direction. Next we can look at (4, -2) to really see the effects. Again we multiply both x and y by 2. This brings our point to (8, -4). If you plot this out, you can visually see the difference between the original graph and the new graph. You apply the same technique to your remaining 2 points to try it for yourself.

E) When flipping a graph, I would recommend doing each flip one at a time and marking out the new points each time. It will be easier to keep track this way. The points on the axis will only change once so we can start with our (6, 0). Flipping this point vertically will not move it, but when it is flipped horizontally, it will become (-6, 0). The point at (0, 2) will be flipped with the same logic, but in the opposite way where it is the vertical flip that will move it. This point will then become (0, -2). A point that will be a bit more tricky is the point (4, -2). You can start by flipping it in either direction, but I will start by flipping it horizontally. The point would then be at (4, 2). Next, I will flip it vertically. This would bring the point to (-4, 2). When you flip a point over an axis, it changes the sign of the value on the opposite axis. When we flipped (4, -2) horizontally over the x-axis, the value of the y-axis, in this case, -2, was changed to +2. You can apply this to your remaining 2 points to try it for yourself.