Just not sure how to do these, not really anything hard.


Use the graph of y = f(x) to graph each function g

Graph has points -4,0 -2,-2 2,2 4,0 0,0

17) g(x) = f(x) -1

19) g(x) = f(x-1)

22) g(x)= f(x+1) -2


I need to know how to work these, answers only will not help me as I can look those up in the book. Please shoe work. I'm sure this is fairly simple I'm just not sure how to do it

Maybe just work one and I'll post with my working the other and u can check

This post was edited by fingerling on Jan 18 2014 11:43pm

Well, i hope you know how to plot ordered pairs (x,y)

f(x) is your equation, in its basic form

f(x)+1 means everything shift 1 up
f(x+1) means shift to the left one

Yeah I know I just haven't had a math course in a couple years and I get confused with all the different variables and where I plug stuff in, etc.


So #22 is shifting up one and right two ?

Why would f(x+1) be to the left ? If its adding one (positive) to x wouldn't it be shifting one to the right ? And how can it move up if its x axis

This post was edited by fingerling on Jan 19 2014 12:09am

if i understand this right your function y=f(x) is given by the 5 points
1: -4,0
2: -2,-2
5: 0,0
3: 2,2
4: 4,0

so the first one (17) would increase the y-value by one, f(x)-1:
1: -4,1
2: -2,-1
5: 0,1
3: 2,3
4: 4,1

the second (18) would require the values for x=-5,-3,-1,1,3 based on f(x-1)
assuming you just connect the dots given for the graph the original would be:
-4,0
-3,1
-2,2
-1,1
0,0
1,1
2,2
3,1
4,0
using the middle value for the missing numbers, because it goes zig-zag let's assume that
-5,1

then f(x-1) is easily established with
-4,1
-3,0
-2,1
-1,2
0,1
etc
because you shift the whole graph by one to the right

now you should be able to solve (22) yourself, with f(x+1)-2
requiring to shift the graph to the left and than subtract 2 which is equal to shifting it down by 2



because you are taking the (x+1)st value, ie the value from (!) the right and not the value of f(x)+1

This post was edited by brmv on Jan 19 2014 12:18am

because you are taking the (x+1)st value, ie the value from (!) the right and not the value of f(x)+1[/QUOTE]



can you rewrite this a different way

This post was edited by fingerling on Jan 19 2014 12:29am

[QUOTE=fingerling,19 Jan 2014 06:28]because you are taking the (x+1)st value, ie the value from (!) the right and not the value of f(x)+1[/QUOTE]
can you rewrite this a different way[/QUOTE]

not really but formally it is
f(x+1) is
f(2) for x=1
f(3) for x=2
etc
which means you are taking the graph and shift it to the left by 1

[QUOTE=brmv,Jan 19 2014 01:54am][QUOTE=fingerling,19 Jan 2014 06:28]because you are taking the (x+1)st value, ie the value from (!) the right and not the value of f(x)+1[/QUOTE]
can you rewrite this a different way[/QUOTE]

not really but formally it is
f(x+1) is
f(2) for x=1
f(3) for x=2
etc
which means you are taking the graph and shift it to the left by 1[/QUOTE]

I guess I'm still a little confused how f(x+1) tells you something different from f(x) -1

Could you try explain that ?

[QUOTE=fingerling,19 Jan 2014 07:07][QUOTE=brmv,Jan 19 2014 01:54am][QUOTE=fingerling,19 Jan 2014 06:28]because you are taking the (x+1)st value, ie the value from (!) the right and not the value of f(x)+1[/QUOTE]
can you rewrite this a different way[/QUOTE]

not really but formally it is
f(x+1) is
f(2) for x=1
f(3) for x=2
etc
which means you are taking the graph and shift it to the left by 1[/QUOTE]

I guess I'm still a little confused how f(x+1) tells you something different from f(x) -1

Could you try explain that ?[/QUOTE]

f(x+1) or f(x-1) or anything similar changes the point the function is to be evaluated
while f(x)+1, f(x)-1 or anything similar changes the value after the function has been evaluated

the first is like 'ask the person to your right/left or similar' while the second compares to asking one more or two more people

does that make sense?

[QUOTE=brmv,Jan 19 2014 02:11am][QUOTE=fingerling,19 Jan 2014 07:07][QUOTE=brmv,Jan 19 2014 01:54am][QUOTE=fingerling,19 Jan 2014 06:28]because you are taking the (x+1)st value, ie the value from (!) the right and not the value of f(x)+1[/QUOTE]
can you rewrite this a different way[/QUOTE]

not really but formally it is
f(x+1) is
f(2) for x=1
f(3) for x=2
etc
which means you are taking the graph and shift it to the left by 1[/QUOTE]

I guess I'm still a little confused how f(x+1) tells you something different from f(x) -1

Could you try explain that ?[/QUOTE]

f(x+1) or f(x-1) or anything similar changes the point the function is to be evaluated
while f(x)+1, f(x)-1 or anything similar changes the value after the function has been evaluated

the first is like 'ask the person to your right/left or similar' while the second compares to asking one more or two more people

does that make sense?[/QUOTE]

OK yes now that part makes sense. Now if if we have f(x) -1 after the function has been evaluated we know we need to increase y value by one. But why ? y= f(x) and f(x)-1 is saying y will be to x-1 , yes ? I just don't see how we go from x-1 to shifting the graph upwards

with f(x)-1 you first evaluate the function at x and then subtract one, so you are reducing the result by one, shifting the graph downwards

f(x-1) shifts the graph by one to the right because when you have x=1 you use the value from x=0 (ie x-1) instead

This post was edited by brmv on Jan 19 2014 01:27am