My Math Forum  

Go Back   My Math Forum > High School Math Forum > Pre-Calculus

Pre-Calculus Pre-Calculus Math Forum


Reply
 
LinkBack Thread Tools Display Modes
May 17th, 2017, 08:49 PM   #1
Newbie
 
Joined: May 2017
From: Surrey

Posts: 1
Thanks: 0

Plz help asap need help now (function transformations)

What happens to the points (12, -6) in the functions:

Y=√f(x) + 2 (f and (x) are both square rooted)

y=-1/3f(3(x+2))-4?
subag24 is offline  
 
June 2nd, 2017, 05:52 AM   #2
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,576
Thanks: 668

Quote:
Originally Posted by subag24 View Post
What happens to the points (12, -6) in the functions:

Y=√f(x) + 2 (f and (x) are both square rooted)
Then you should have written y= √(f(x)+ 2).
Now, when you refer to the point (12, -6) do you mean that (12, -6) is on the graph of y= f(x) so that f(12)= -6? If so then f(x)+ 2= -4 and √(-4) is not a real number.

Quote:
y=-1/3f(3(x+2))-4?
Another ambiguity. Do you mean 1/(3f) or (1/3)f? Do you mean 1/(3f- 4) or 1/(3f)- 4?

Assuming, as above, that you mean that f(12)= -6, to evaluate f(3(x+ 2)) with only that information, must have 3(x+ 2)= 12 so that x+2= 4 and x= 2. In that case, if you intend (-1/3)(f(3(x+2))- 4= that is (-1/3)(-6)+ 4= 2+ 4= 6. If you intend -1/(3f(3(x+2)- 4 that is -1/(-1+ 4= 4+ 1/18. If you intend -1/(3f(3(x+2)- 4) that is -1/(-18+ 4)= 1/14.
Country Boy is offline  
Reply

  My Math Forum > High School Math Forum > Pre-Calculus

Tags
asap, function, plz, transformations



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Need help asap! Zery Algebra 6 August 1st, 2015 10:58 PM
ASAP! Please Help With This! Hithere Algebra 1 November 26th, 2013 12:22 PM
I need help ASAP HBDB Algebra 2 August 6th, 2008 07:14 AM
help please ASAP helpmewithmath Algebra 2 June 13th, 2008 06:01 PM
help pls !!! asap haruka-san Advanced Statistics 2 November 28th, 2007 01:54 PM





Copyright © 2017 My Math Forum. All rights reserved.