January 26th, 2018, 07:00 PM  #1 
Newbie Joined: Jan 2018 From: CA Posts: 7 Thanks: 0  functions help with graphing
not really sure even how to approach this problem the pic is attached 
January 26th, 2018, 07:23 PM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,648 Thanks: 574 Math Focus: Yet to find out. 
Perhaps start by defining the function $f(x)$, then you have something to work with. Note: $f(x) = mx + c$, can you find the slope $m$ and $y$intercept $c$? 
January 26th, 2018, 07:37 PM  #3 
Newbie Joined: Jan 2018 From: CA Posts: 7 Thanks: 0 
thanks for the quick reply. Yes the slope would be 1 or B=1 because m= y2y1 __________ x2x1 
January 26th, 2018, 07:38 PM  #4 
Newbie Joined: Jan 2018 From: CA Posts: 7 Thanks: 0  
January 26th, 2018, 07:47 PM  #5 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,648 Thanks: 574 Math Focus: Yet to find out.  
January 26th, 2018, 07:51 PM  #6  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,648 Thanks: 574 Math Focus: Yet to find out.  Quote:
We could replace 'c' by 'b' or 'e' or any other letter of symbol if we wanted to.  
January 26th, 2018, 08:11 PM  #7  
Newbie Joined: Jan 2018 From: CA Posts: 7 Thanks: 0  Quote:
Quote:
so point (5,4) would now become point (5,3) So A is the correct answer?  
January 26th, 2018, 08:11 PM  #8 
Newbie Joined: Jan 2018 From: CA Posts: 7 Thanks: 0  
January 26th, 2018, 08:29 PM  #9  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,648 Thanks: 574 Math Focus: Yet to find out.  Quote:
$f(f(x)) = f(x1) = (x1)1 = x  2 = g(x)$. For arbitrary slope and intercept: $f(x) = mx + b$, $f(f(x)) = f(mx + c) = m(mx + c) + c = m^2 x + mc + c = m^2 x + c(m + 1)$  
January 26th, 2018, 09:20 PM  #10 
Newbie Joined: Jan 2018 From: CA Posts: 7 Thanks: 0 
Thanks


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