Expert II Reflections, Transformations, and Operations With Functions 30-1 Only Flashcards
(14 cards)
Vertical translations and general math 30-1 notation for transformations
In math 30-1 we need to revisit some notation on what we already know about vertical translations. You already know that y = x² can be compared to y = x² + k, and that this will make the resulting graph move up k units, giving a vertical translation of k units. You also know that the opposite direction can be indicated by using negative values, so if k was equal to -3, then the graph would move down by 3 units while maintaining its shape.
So in math 30-1 what we need to do is to understand new language around this concept. For example, I could give you a function such as y = f(x) + k and you should be able to say that k shows the vertical translation, so the graph could move up k units. Then you could take every point and move k units vertically while maintaining the same x value of that point. Notice how we don’t know what the function is, but we are still able to make this conclusion.
Now, to expand on this even further, what would happen if we were to try to isolate for x? In all the above examples, we have isolated for y and so k is showing upwards motion if k is positive. But if we were to move k to the other side of the equation, that would be the first move towards trying to isolate for x. We could subtract k from both sides and successfully move k to the other side.
The notation y - k = f(x) means y = f(x) + k so the transformation for both of these would be a vertical translation by k units.
This allows us to use new language such as ˋIf y is replaced with y-k, then what is the resulting transformation?ˋ and to answer you should write y - k = f(x) because y = f(x) but now we are replacing y with y - k. Then solve y - k = f(x) for y so that it looks like our prior learning. You will see that you get y = f(x) + k and we know this to be similar to y = x² + k where the graph would be vertically translated by k units (up k units).
The point is that we are used to solving for y when looking at transformations, so continue to make questions look like that and solve for y to make sure that you are not getting confused by the new wording.
In short (x, y) –> (x, y - k) shows a vertical translation by k units since y is being replaced by y - k.
y = f(x) +k also shows a vertical translation by k units.
Notice that the signs change when talking about replacing the variable with something that still contains that variable compared to solving for that variable. This pattern is what is consistent throughout all the transformations you will be learning. Essentially, solving for the variable will feel opposite to talking about the replacement of that variable by the new thing.
Summarize all the transformations that could happen on the graph of y = f(x).
y = af(b(x-h)) + k
a is the vertical stretch, so if it is 2, all y values would be 2 times larger in the new graph for the same x
y is replaced by y/a so y –> y/a means that we have y/a = f(x) –> y = af(x)
if a is negative it produces a reflection about the x-axis like a calendar flip.
1/b is the horizontal stretch, so if b is 2, all x values would be half their values for the same y (the horizontal stretch factor is a reciprocal of b)
x is replaced by bx so x –> bx and y = f(x) –> y = f(bx)
If b is negative, it produces a reflection about the y-axis, so that is a book flip.
h is the units the graph moves right, so if h is 2, all x values would be moved right 2 units, but in the equation you would see a negative sign in front of that 2
x is replaced by x - h so x –> x-h and y = f(x) –> y = f(x-h)
k is the units the graph moves up, so if k is 2, the resulting graph would be 2 units higher
y is replaced by y - k so y –> y-k and y=f(x) –> y - k = f(x) –> y = f(x) + k
Notice how there is a negative placed in front of h and that the b value seems to be needing the idea of division? This is because this equation is solved for y and not for x. When we have isolated the variable, then the transformations will seem predictable. When we are replacing a variable with that variable + or - or mutiplied or divided by something, then everything should seem backwards.
The key information here is that all this content operates the same way as our transformations on the parabola from grade 11. We just need to remember that the values that influence x feel like they are working in an opposite direction compared to what we see happening to the y values. The reason for this is explained best by isolating many graphs for x and looking at the resulting transformations. Then bring the notation back to solved for y and you will see that it is quite logical.
Summarize all the information required to perform operations with two functions.
You can add two functions together:
h(x) = f(x) + g(x)
In this case, plot both f(x) and g(x), then for every x value, add the two y values together to get the new y value for h(x) while plotting it in line with the x value that you were working with.
You can also subtract one function from another using the same technique but with a subtraction instead of an addition. h(x) = f(x) - g(x)
It is also possible to multiply and divide functions using the same concepts.
h(x) = f(x) dot g(x)
h(x) = f(x) / g(x)
Solve for the variable vs. the variable being replaced by ….
Usually we are used to solving for y, while looking at how x is replaced by say x-2.
These are opposite ideas and so we need to make sure to recognize the words ˋreplaced byˋ as making things happen opposite to the usual (meaning compared to solving for the variable). If you solve for the variable, the function should behave predictably and you should be able to see everything more clearly.
What happens to the graph of the function y = f(x) if the following changes are made to its equation?
replace x with x + 2
replace y with y - 8
The graph will move 2 units to the left, and 8 units up.
We are used to replacing x. When replacing y we must solve for x in our minds and think about what that k value must be. K is equal to 8 here y - 8 = f(x) becomes y = f(x) + 8 when we solve for y and we can see that the k value is 8.
The replacement of x by x + 2 is something we are used to seeing in parabolas and thus takes no rearranging: y = f(x+2) has always meant a shift 2 units to the left in a parabola, and nothing has changed here when we apply the concept to all functions. Think y = f(x-(-2)) and you will see that h = -2.
The point (2, -3) lies on the graph of y = f(x). State the coordinates of the image of the point under the following transformation:
y + 8 = f(x)
y + 8 = f(x)
y = f(x) - 8
so k = -8 and we have a vertical translation of -8 units up, or simply put, 8 units down
so (2, -3) changes to a point 8 units down from there giving (2 , -3 - 8) = (2, -11)
Notice how there is no change to the x coordinate.
Define a, b, h, and k
a = vertical stretch by a factor of a
b = horizontal stretch by a factor of 1/b
h = horizontal translation
k = vertical translation
What is an inverse function?
This is a type of transformation where the input and output are exchanged.
The notation of the inverse function is y = f-1(x)
If the inverse is not a function, then just change the x and y values and state x = f(y), but you should not need to do this. Just make sure that you create a function when you do the inverse by placing restrictions on the domain of the original so that you can use the function notation instead of doing the ideas in this paragraph.
The characteristics of an inverse function:
-reflection about the line x=y
Order of transformations
follow this order when graphing a new graph based on transformations and a given graphed function:
1. Stretches (vertical and horizontal)
2. Reflections
3. Translations (vertical and horizontal)
When doing the opposite, so when looking at what transformations must have happened, look for things like amplitude and period on the sine graph to determine stretches, because otherwise it is easy to get confused with any translations that also occured. Translations can be seen on a sine graph with the midline. Essentially, you need to know all your basic graphs really well so that you can see if they look stretched or not.
What are invariant points? Give examples of general situations where these occur.
Points that do not change during transformations.
For example, x-intercepts are invariant under a vertical stretch since x-int is where y = 0 and multiplying something by 0 will have a product of 0, unchanged from before.
y-intercepts are invariant under horizontal stretches.
Points on the y=x line will be invariant when performing the inverse.
A net translation cannot yield any invariant points.
When applying transformations, what are key things to consider about the resulting graph? How might you describe the resulting graph? What are all the things that you could be asked about in that resulting graph?
- state domain, range, max, and min by comparing to the original graph and applying the transformations directly to predict these
- state x-int and y-int as well as invariant points
- show what happens to (x,y) and find the image point of it by applying transformations so that the new point becomes (x/b + h, ay + k)
- if a rational function, you must state all asymptotes, and in general state these when they arise
- end behaviour
Describe the transformations on this function:
y = f(2x-6)
You must first factor to find the h value properly:
y = f(2(x-3))
Then you can see that b is 2, and we have a horizontal stretch by a factor of 1/b, so here there is a horizontal stretch of 1/2. We can also see that h is equal to 3, so we have a translation 3 units right.
The b value is the reciprocal of the stretch factor.
What is the difference between mapping notation and replacement?
Mapping notation looks like this: (x, y) –> (x + 3, y + 4)
This means that the graph has moved 3 units to the right, and 4 units up
You will see the ordered pair in mapping notation.
The x coordinate was x and now you are adding 3 to the x coordinate.
So h = 3 and k = 4
Replacement:
x is replaced by x -3 and y is replaced by y - 4 is a replacement, and not a mapping. Replacement and mapping are not the same thing!
This means y = f(x) –> y - 4 = f(x-3) –> y = f(x-3) + 4 so h = 3 and k = 4
So the replacement and mapping notation above showed the same transformations but the notation was different!
Make sure that you know the shapes of all types of graphs in their most basic form. This includes:
- y = mx + b
- y = x²
- y = x³
- y = |x|
- y = 1/x (remember asymptotes at y = 0 and x = 0)
- y = square root of x
For more of a challenge, also include these:
- y = 2x
- y = (1/2)x
- y = log(x)
- y = sin(x)
- y = cos(x)
- y = tan(x)
Try these on your graphing calculator but only after you have first graphed by hand with a table of values.
