Describe how the grah of each of the following functions

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Homework Help Overview

The discussion revolves around understanding how to graph transformations of functions, specifically the function 3f(2x) - 6, based on the graph of y = f(x). Participants are exploring the effects of vertical and horizontal stretches and shifts on the graph.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of horizontal compression and vertical stretching, questioning how to apply these transformations to graphing. There are inquiries about whether to use a table of values or a step pattern for plotting points on the transformed graph.

Discussion Status

The conversation is ongoing, with participants sharing their understanding of step patterns for parabolas and how they relate to transformations. Some guidance has been offered regarding the use of two points to define a linear graph, but there remains uncertainty about the best approach for horizontal stretches and the application of step patterns.

Contextual Notes

Participants are navigating the complexities of function transformations, with some expressing confusion about the terminology and methods used in graphing. There is a mention of specific values from a table of f(x) that may not always be provided, adding to the discussion's complexity.

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Homework Statement


>Describe how the grah of each of the following functions can be obtained from the graph y=f(x)

3f(2x)-6



Homework Equations



f(x)=a(x-h)k
y=f(x)

The Attempt at a Solution



I know that its going to be a vertical stretch by a factor of 3, and a horizontal compression by a factor of 1/2 . But i have two questions, when its horizontal comrpession do i always turn it into a fraction? , and when its something like y=f(1/2x) -6 , what do i multiply the 1/2 by to get the actual value for the horizontal stretch?

Also, How do i graph it? Do I have to use a table of values, Always? inorder to get the points for the horizontal compression/stretch or is there an easier way?

Like with vertical stretch / compression you can multiply the step pattern 1,3,5 by whatever the a value is.
 
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As "2x" goes from a to b, x goes from a/2 to b/2- that's why there is "a horizontal compression by a factor of 1/2". As "(1/2)x" goes from a to b, x goes from 2a to 2b. To get "compression" in f(ax) you divide by a. If a= 2 then you divide by 2 which is the same as multiplying by 1/2. If a= 1/2 then you divide by 1/2 which is the same as multiplying by 2.

If, when you write "f(x)= a(x- h)k" both a and k are constants, then f is linear- the graph of y= f(x) is a straight line and so is the graph of y= 3f(2x)- 6. Since a line is determined by two points, just calculating y= 3f(2x)- 6 for two values of x gives two points and then just draw the line containing them.

More generally, if you know the graph of y= f(x), the graph of y= 3f(2x)- 6 is exactly the same except that it is, as you say, "stretched by 3" vertically, "compressed by a factor of 1/2" horizontally, and moved down 6.
 


so you still multiply by the step pattern of y=f(x) ? or you use a table of values and plug in whatever is needed from the translations of y=f(x)
 


Nelo said:
so you still multiply by the step pattern of y=f(x) ? or you use a table of values and plug in whatever is needed from the translations of y=f(x)
I have no idea what you mean by this. What "step pattern" are you talking about?
 


Step pattern of a parabola at y=x^2 is 1,3,5

Start at vertex, go out by one, go up by 1 and plot the point. From that point, you go out by 1 go up by 3, plot the point. then go out by one, go up by 5 and plot the point.. use symmetry to get the other side and that gives you y=x^2 of a porabola.

If you have something like y=4(x-2) +4

then the vertex is (2,4) , and since there is a vertical stretch you multiply the step pattern of, 1,3,5 by 4 , giving you, 4 ,12, 20, ... effectively... Start at the vertex of 2,2 . and go out by 1 and up by 4, plot the point, then go out by one up by 12, plot the point. and so forth, that is what I've learned as the step pattern for a parabola. when you have vertica lstretch you multiply by 4. Since now we're doing horizontal stretch I was wondering if you could relate the 1,3,5 step patern inorder to get the points on a horizontal stretch, cause I really don't get how to do it.
 


Does this method work then?

The Table of values for f(x) is always : -2 ,-1 , 0 , 1 , 2, 3, 4. With the corresponding y values of 0 2 4 3 2 1.

If there is a horizontal stretch of f(2x)
Then you need to use values for x that will equal -2 on the y=f(x) table.

Ie) If y=(2x) and if x = -1 then y = f(2*(-1)) = f(-2) = 0 for the you value on the f(2x) table.

Thereforemmaking the first point 0,0

Does this make any sense?

Or will y= f(x) values always be given?
 

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