Graphing functions to form the letters CAS
- Thread starter kt5
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I don't really know how to graph the equation you gave, but I am looking for a function, something I can graph, like X^2, to form those three letters. My assignment says, "Using the graphing calculator, come up with a graph or series of graphs to give your initials. Use at least 5 functions, 3 of which can be differentiated." I hope I made myself a little more clear, I'm just having trouble coming up with functions that resemble or create those letters. Any ideas? Thanks for any help.
C - You could try using half a circle or 3/4 of the circle or whatever you think makes it look like a 'C'.
A - Try splitting it into 3 linear equations (y = mx + b). Of course you would have to restrict their domains to make it look like an A
S - Try splitting it into 3 pieces: 2 half circles (the bottom and top arc's of the 'S') and make use of the arccosx graph for the middle portion. I don't know how this would appear on the graphing calculator but hopefully, it won't look too distorted.
A - Try splitting it into 3 linear equations (y = mx + b). Of course you would have to restrict their domains to make it look like an A
S - Try splitting it into 3 pieces: 2 half circles (the bottom and top arc's of the 'S') and make use of the arccosx graph for the middle portion. I don't know how this would appear on the graphing calculator but hopefully, it won't look too distorted.
Thank you for your reply.
I had been trying to use a half circle for the 'C', however when I graph it, it's bisected by the y-axis, and I can't figure out how to rotate it to the left to be in the right position for a 'C', or so it is bisected by the x-axis. I understand your explanation for the 'A' and I know I have to restrict the domains, but I'm not sure how to do that on the graphing calculator. And I think I'll try to just use two half circles for the 'S', because we haven't learned about the arccosx graph yet. I know that the procedure might be something simple I'm missing, so if you are able to explain my confusion on the 'C' and 'A' it'd be greatly appreciated. Thank you again.
I had been trying to use a half circle for the 'C', however when I graph it, it's bisected by the y-axis, and I can't figure out how to rotate it to the left to be in the right position for a 'C', or so it is bisected by the x-axis. I understand your explanation for the 'A' and I know I have to restrict the domains, but I'm not sure how to do that on the graphing calculator. And I think I'll try to just use two half circles for the 'S', because we haven't learned about the arccosx graph yet. I know that the procedure might be something simple I'm missing, so if you are able to explain my confusion on the 'C' and 'A' it'd be greatly appreciated. Thank you again.
C:
Oh does your teacher take into consideration of the axes? I doubt that as it is kind of unreasonable ... But if so, just shift your circle so that it doesn't touch any of the axes. I don't know what you mean by "rotating it to the left" since you can't really distinguish any part of the circle.
A:
You don't have to do it on the graphing calculator. If I told you that f(x) = 3x + 5 on the interval [2, 5], all you would consider are the x values from 2 to 5. If you insist on restricting the domain on the graphing calculator, why don't you just change the settings of your window?
S:
Hmm that poses a problem. You can't just use two half circles to get an S, it's impossible. Maybe instead of cos[sup:q101ifvw]-1[/sup:q101ifvw]x, you could use the middle portion of y = x[sup:q101ifvw]3[/sup:q101ifvw]? It kind of resembles the middle portion of the 'S'. Just make sure you restrict the domain so that you don't have to consider the rest of x[sup:q101ifvw]3[/sup:q101ifvw]
Oh does your teacher take into consideration of the axes? I doubt that as it is kind of unreasonable ... But if so, just shift your circle so that it doesn't touch any of the axes. I don't know what you mean by "rotating it to the left" since you can't really distinguish any part of the circle.
A:
You don't have to do it on the graphing calculator. If I told you that f(x) = 3x + 5 on the interval [2, 5], all you would consider are the x values from 2 to 5. If you insist on restricting the domain on the graphing calculator, why don't you just change the settings of your window?
S:
Hmm that poses a problem. You can't just use two half circles to get an S, it's impossible. Maybe instead of cos[sup:q101ifvw]-1[/sup:q101ifvw]x, you could use the middle portion of y = x[sup:q101ifvw]3[/sup:q101ifvw]? It kind of resembles the middle portion of the 'S'. Just make sure you restrict the domain so that you don't have to consider the rest of x[sup:q101ifvw]3[/sup:q101ifvw]
You don't say which calculator you are using. These steps work pretty well for TI-83 Plus. If they don't work for you or you don't like the equations, PM me and I'll delete this.
Key names are in blue.
abs is absolute value. MATH NUM 1 gets you there.
> and < are to limit the domains. Use 2nd MATH to get to the test menu.
Just because you havn't gotten to arccos in class yet doesn't mean you can't use it. You know what it is from TRIG.
Pi is of course 3.14159...
To use Y[sub:1vrwffkn]3[/sub:1vrwffkn] in Y[sub:1vrwffkn]5[/sub:1vrwffkn] use VARS Y-VARS 1 3
For the axis problem
2nd ZOOM gives a format menu. Arrow down to "AxesOff" and hit ENTER.
Set WINDOW to -3, 3, 1, -1, 8, 1, 1
The Y='s are:
Y[sub:1vrwffkn]1[/sub:1vrwffkn] = (2-2*abs(X))*Pi
Y[sub:1vrwffkn]2[/sub:1vrwffkn] = Pi*(2-2*abs(X))(abs(X)<.5)
Y[sub:1vrwffkn]3[/sub:1vrwffkn] = abs( 2nd SIN(X-2))
Y[sub:1vrwffkn]4[/sub:1vrwffkn] = 2nd COS(X-2))+ PI/2
Y[sub:1vrwffkn]5[/sub:1vrwffkn] = 2 * PI - Y[sub:1vrwffkn]3[/sub:1vrwffkn]
Y[sub:1vrwffkn]6[/sub:1vrwffkn] = Pi*sqrt(1-(x+2)^2))+Pi
Y[sub:1vrwffkn]7[/sub:1vrwffkn] = Pi - Pi(sqrt(1-(x+2)^2)
Y[sub:1vrwffkn]8[/sub:1vrwffkn] = 0
Key names are in blue.
abs is absolute value. MATH NUM 1 gets you there.
> and < are to limit the domains. Use 2nd MATH to get to the test menu.
Just because you havn't gotten to arccos in class yet doesn't mean you can't use it. You know what it is from TRIG.
Pi is of course 3.14159...
To use Y[sub:1vrwffkn]3[/sub:1vrwffkn] in Y[sub:1vrwffkn]5[/sub:1vrwffkn] use VARS Y-VARS 1 3
For the axis problem
2nd ZOOM gives a format menu. Arrow down to "AxesOff" and hit ENTER.
Set WINDOW to -3, 3, 1, -1, 8, 1, 1
The Y='s are:
Y[sub:1vrwffkn]1[/sub:1vrwffkn] = (2-2*abs(X))*Pi
Y[sub:1vrwffkn]2[/sub:1vrwffkn] = Pi*(2-2*abs(X))(abs(X)<.5)
Y[sub:1vrwffkn]3[/sub:1vrwffkn] = abs( 2nd SIN(X-2))
Y[sub:1vrwffkn]4[/sub:1vrwffkn] = 2nd COS(X-2))+ PI/2
Y[sub:1vrwffkn]5[/sub:1vrwffkn] = 2 * PI - Y[sub:1vrwffkn]3[/sub:1vrwffkn]
Y[sub:1vrwffkn]6[/sub:1vrwffkn] = Pi*sqrt(1-(x+2)^2))+Pi
Y[sub:1vrwffkn]7[/sub:1vrwffkn] = Pi - Pi(sqrt(1-(x+2)^2)
Y[sub:1vrwffkn]8[/sub:1vrwffkn] = 0
