# Homework Help: Describe how the grah of each of the following functions

1. Jul 15, 2011

### Nelo

1. The problem statement, all variables and given/known data
>Describe how the grah of each of the following functions can be obtained from the graph y=f(x)

3f(2x)-6

2. Relevant equations

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

3. 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.

2. Jul 15, 2011

### HallsofIvy

Re: Functions

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.

3. Jul 15, 2011

### Nelo

Re: Functions

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)

4. Jul 16, 2011

### HallsofIvy

Re: Functions

I have no idea what you mean by this. What "step pattern" are you talking about?

5. Jul 16, 2011

### Nelo

Re: Functions

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 ive 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 dont get how to do it.

6. Jul 16, 2011

### Nelo

Re: Functions

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 ya 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?