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)
Y = f(x - 2)
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.
chances are using a T-chart is either a waste of time or too much work, use transformations
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
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