1. ## Graphing Functions

I'm totally stumped as to how to do this because it's a bit confusing to me.

I'm asked to determine and graph how this function should be

y = f(x - 2)

Now when I make a T chart and pick for x values 0 , 1 , 2 , 3, 4, 5, 6, 7, the line is linear when in fact the answer the book portrays is an increasing and decreasing line. Help please let me know what i'm missing from doing this right Thanks!

Oh and what exactly is the difference between

F(x) = (x - 2)
and
Y = f(x - 2)

2. Originally Posted by JonathanEyoon
I'm totally stumped as to how to do this because it's a bit confusing to me.

I'm asked to determine and graph how this function should be

y = f(x - 2)
this is just the graph of y = f(x) shifted 2 units to the right

Oh and what exactly is the difference between

F(x) = (x - 2)
and
Y = f(x - 2)
no difference, just notation really. we often call y a function of x, that is, y = f(x)

3. Originally Posted by Jhevon
this is just the graph of y = f(x) shifted 2 units to the right

no difference, just notation really. we often call y a function of x, that is, y = f(x)

So the graph should be a linear line with an x-intercept at (2,0) with a slope of X right?

how would you use a T-chart to determine the points for y= f(x - 2)?

It would just be like this right?

X l Y= f(x -2)
0 l -2
1 l -1
2 l 0
3 l 1
4 l 2

4. Originally Posted by JonathanEyoon
So the graph should be a linear line with an x-intercept at (2,0) with a slope of X right?

how would you use a T-chart to determine the points for y= f(x - 2)?

It would just be like this right?

X l Y= f(x -2)
0 l -2
1 l -1
2 l 0
3 l 1
4 l 2

we need to know what f(x) is. what does the original function look like? or what are its points (intercepts and such)?

5. Originally Posted by Jhevon
we need to know what f(x) is. what does the original function look like? or what are its points (intercepts and such)?

It doesn't state what the original function is. I'll give you the exact instructions of the problem and what i'm asking to do

- The graph of f is given. Sketch the graphs of the following functions.

a) y = f(x - 2)

b) y = f(x) - 2

c) y = 2f(x)

d) y = -f(x) + 3

Those are the exact instructions. So I thought I could just use a T-chart and sketch the graph of each...but the thing is i'm getting the complete wrong graph.

6. Originally Posted by JonathanEyoon
It doesn't state what the original function is. I'll give you the exact instructions of the problem and what i'm asking to do

- The graph of f is given. Sketch the graphs of the following functions.

a) y = f(x - 2)

b) y = f(x) - 2

c) y = 2f(x)

d) y = -f(x) + 3

Those are the exact instructions. So I thought I could just use a T-chart and sketch the graph of each...but the thing is i'm getting the complete wrong graph.
to fully help you i need to know what the function looks like. otherwise, i can only tell you how to shift the function to get the desired functions. you are required to use your knowledge of "transformations" here. do you know the transformation rules?

chances are using a T-chart is either a waste of time or too much work, use transformations

7. Originally Posted by Jhevon
to fully help you i need to know what the function looks like. otherwise, i can only tell you how to shift the function to get the desired functions. you are required to use your knowledge of "transformations" here. do you know the transformation rules?

chances are using a T-chart is either a waste of time or too much work, use transformations

I'll try it with just transformations. Y = f(x - 2) IS a linear line though right? Because just using transformations, i know that the the x-intercept should be (2,0).

8. Originally Posted by JonathanEyoon
I'll try it with just transformations. Y = f(x - 2) IS a linear line though right? Because just using transformations, i know that the the x-intercept should be (2,0).
no, it is not necessarily a line! it depends on the original function, it could be whatever

say we had f(x) = x^2

then f(x - 2) = (x - 2)^2 ...........not a line, it is of the form of the original, a parabola, but shifted

DO NOT MIX UP y = x - 2 AND y = f(x - 2) ! THEY ARE SAYING TWO COMPLETELY DIFFERENT THINGS!

9. Originally Posted by Jhevon
no, it is not necessarily a line! it depends on the original function, it could be whatever

say we had f(x) = x^2

then f(x - 2) = (x - 2)^2 ...........not a line, it is of the form of the original, a parabola, but shifted

DO NOT MIX UP y = x - 2 AND y = f(x - 2) ! THEY ARE SAYING TWO COMPLETELY DIFFERENT THINGS!

You wouldn't happen to have a copy of the textbook "Precalculus" 5th Edition by James Stewart do you ? The exercise is chapter 2.4 #19 and it doesn't state what the function is. So if you're saying it can't be graphed without the original function which is not given, then i'm not seeing how this book wants me to graph it haha. *sigh* i'll ask the professor to see what is going on. Thanks for your help

10. Originally Posted by JonathanEyoon
You wouldn't happen to have a copy of the textbook "Precalculus" 5th Edition by James Stewart do you ? The exercise is chapter 2.4 #19 and it doesn't state what the function is. So if you're saying it can't be graphed without the original function which is not given, then i'm not seeing how this book wants me to graph it haha. *sigh* i'll ask the professor to see what is going on. Thanks for your help
nope, don't have that book sorry. but you should be able to know what to do anyway.

for f(x - 2) you shift the graph to the right, 2 units. so you pretty much redraw the graph beside itself and indicate that it is 2 units apart (by identifying intercepts or anything like that)

for f(x) - 2 you shift the ORIGINAL graph down 2 units. so you take each point and redraw it in the same vertical line 2 units below its original location

for 2f(x) you double the y-value for the same x-values. so if it seems like the graph goes through, (1,3), you would change that point to (1,6) and so on. it should kind of look like you're squishing the graph and making it steeper

for -f(x) + 3 there are two operations we need to do. one that takes care of the -f(x) and one that takes care of the +3. for the -f(x), this means we reflect the graph in the y-axis, that is, flip it upside down (think of y = x^2 and y = -x^2). so you treat the y-axis as a mirror and retrace the graph so that the new graph looks just like the old (including the x-intercepts) but it's upside down (you would change all the y-intercepts to their negatives. for the +3, you shift that new graph UP 3 units