My Math Forum functions help with graphing

 Pre-Calculus Pre-Calculus Math Forum

January 26th, 2018, 07:00 PM   #1
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functions help with graphing

not really sure even how to approach this problem

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 January 26th, 2018, 07:23 PM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,834 Thanks: 649 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 $c$? Thanks from sara654
 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= y2-y1 __________ x2-x1
January 26th, 2018, 07:38 PM   #4
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Quote:
 Originally Posted by Joppy 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 $c$?
doesnt f(x) = mx+b and not c?

or am I thinking of something different?

January 26th, 2018, 07:47 PM   #5
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 Originally Posted by sara654 thanks for the quick reply. Yes the slope would be 1 or B=-1 because m= y2-y1 __________ x2-x1
Correct, $f(x) = x - 1$. Now can you find $g(x) = f(f(x)) = (f \circ f)(x)$?

January 26th, 2018, 07:51 PM   #6
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 Originally Posted by sara654 doesnt f(x) = mx+b and not c? or am I thinking of something different?
Sometimes we use different letters to represent the same things, what often matters is the 'form' or 'structure' of your expression.

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
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Quote:
 Originally Posted by Joppy 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 $c$?
Quote:
 Originally Posted by Joppy Correct, $f(x) = x - 1$. Now can you find $g(x) = f(f(x)) = (f \circ f)(x)$?
So correct me if I am wrong but all I would do is subtract -1 from each of the Y values.

so point (5,4) would now become point (5,3)

So A is the correct answer?

January 26th, 2018, 08:11 PM   #8
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Quote:
 Originally Posted by Joppy Sometimes we use different letters to represent the same things, what often matters is the 'form' or 'structure' of your expression. We could replace 'c' by 'b' or 'e' or any other letter of symbol if we wanted to.
oh okay

thank you

January 26th, 2018, 08:29 PM   #9
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 Originally Posted by sara654 So correct me if I am wrong but all I would do is subtract -1 from each of the Y values. so point (5,4) would now become point (5,3) So A is the correct answer?
In this case you have a slope of 1 so yes, A is correct. I recommend you to check by performing the usual operations:

$f(f(x)) = f(x-1) = (x-1)-1 = x - 2 = g(x)$.

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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