I'm not sure if the topic in a right forum...but I'm so desperate ;_; can anyone help me with the trig problems?
Here are the problems...
1.) Plot the point symmetric to the line y=x (f(x)=x) for each point given below
a. (2,5)
b. (-1,-3)
^ I got the answer 2.5 for the first one and 3 for the second one, am I correct?
2.) Sketch the graph f(x )=3^x-2 on the interval (-3,3). Make a table of your values.
^ I don't really understand
Thanks for the help.
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Trigonometry
#2
Posted 10 May 2008 - 07:29 PM
err
1)a. should be (2, -1)
b. should be (-1, 1)
2) table should be
x | f(x)
-3 | 1/27 - 2
-2 | 1/9 - 2
-1 | 1/3 -2
0 | -1
1 | 1
2 | 7
3 | 25
1)a. should be (2, -1)
b. should be (-1, 1)
2) table should be
x | f(x)
-3 | 1/27 - 2
-2 | 1/9 - 2
-1 | 1/3 -2
0 | -1
1 | 1
2 | 7
3 | 25
#3
Posted 10 May 2008 - 07:33 PM
QUOTE (Voltage @ May 10 2008, 09:29 PM) <{POST_SNAPBACK}>
err
1)a. should be (2, -1)
b. should be (-1, 1)
2) table should be
x | f(x)
-3 | 1/27 - 2
-2 | 1/9 - 2
-1 | 1/3 -2
0 | -1
1 | 1
2 | 7
3 | 25
1)a. should be (2, -1)
b. should be (-1, 1)
2) table should be
x | f(x)
-3 | 1/27 - 2
-2 | 1/9 - 2
-1 | 1/3 -2
0 | -1
1 | 1
2 | 7
3 | 25
Thanks...do you mind telling me how did you find the answers? I gotta learn it for the test ^^;
And what do you mean by 1/27-2? o_o Thanks xD!
Get Real people, Sherry Always Me~ ^.^!! 
#4
Posted 10 May 2008 - 07:36 PM
I won't give you answers directly. However, I can guide you.
2) The intervals between -3, and 3 refer to the values on the X axis. Look at what's happening between -3 and 3 on x.
Here's a hint -> 3^x where x = negative number. That negative indicates an inverse so it's 1/3^2 for 3^-2. Just look at the values between -3 and 3
.
2) The intervals between -3, and 3 refer to the values on the X axis. Look at what's happening between -3 and 3 on x.
Here's a hint -> 3^x where x = negative number. That negative indicates an inverse so it's 1/3^2 for 3^-2. Just look at the values between -3 and 3
#5
Posted 10 May 2008 - 07:43 PM
oh i'm starting to think the f(x) equation is really f(x) = 3^(x-2)
if that's the case then what i wrote is wrong
first the first part it's easier to visualize. just draw the line y = x which is just a line through the origin that has slope 1 (by incrementing by integer unit to the right by one horizontally, the same thing happens to the vertical unit by incrementing by one)
then plot the point (2,5). Notice how far it's y component is from the y=x line. The objective is if you folded your piece of paper on the line y=x, would the point you find land right on top of (2,5) ?
anyways, that point is (5,2)
for point (-1, -3), do the same thing and get (-3, -1)
also notice that finding the symmetry across y=x line isalways swaps the position of the x and y coordinates.
I notice actually note that my previous answer to that was incorrect
if that's the case then what i wrote is wrong
first the first part it's easier to visualize. just draw the line y = x which is just a line through the origin that has slope 1 (by incrementing by integer unit to the right by one horizontally, the same thing happens to the vertical unit by incrementing by one)
then plot the point (2,5). Notice how far it's y component is from the y=x line. The objective is if you folded your piece of paper on the line y=x, would the point you find land right on top of (2,5) ?
anyways, that point is (5,2)
for point (-1, -3), do the same thing and get (-3, -1)
also notice that finding the symmetry across y=x line isalways swaps the position of the x and y coordinates.
I notice actually note that my previous answer to that was incorrect
#6
Posted 10 May 2008 - 07:47 PM
Oh no, it's actually f(x)=3^(x)-2 <--- the -2 is not in the exponent...
Thanks for helping guys...
Thanks for helping guys...
Get Real people, Sherry Always Me~ ^.^!! 
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