Transformations of Functions- graphing y=f(x)-4?

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SUMMARY

The discussion centers on graphing the function y = f(x) - 4, emphasizing the necessity of defining f(x) before graphing. Participants agree that the graph of f(x) should be shifted down four units, resulting in a horizontal line at y = -4 if f(x) is assumed to be 0. However, they clarify that without a specific function for f(x), the graph cannot be accurately plotted. The conversation highlights the importance of understanding function definitions in graphing transformations.

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How would you graph y=f(x)-4? I am not sure how the original graph looks like y=f(x) either. Also, if i were to graph this using a graphing calculator, how would that be done?
 
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Erm.. you need to know what f(x) is. f(x) can be any function of x, so it is impossible to sketch it until you say what it is!
 
I think the answer kumar wants is:

You take the graph of f(x) and shift it down four units. (Assuming f:R->R).
 
I think ZioX is right. From what I remember, f(x) is the same thing as y so it's saying (IMO): y = -4. So it would be shifted down 4 units and you should have a horizontal line.
 
ZioX said:
I think the answer kumar wants is:

You take the graph of f(x) and shift it down four units. (Assuming f:R->R).

Well yes, but in the original post he says he doesn't know what the graph of the function f(x) looks like. Thus, he will need to plot that first then shift it four units down.
 
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Prim3 said:
I think ZioX is right. From what I remember, f(x) is the same thing as y so it's saying (IMO): y = -4. So it would be shifted down 4 units and you should have a horizontal line.

Huh? That's not what ziox is saying. How do you get that the graph is even a straight line? We do not know the form of f(x). It could be f(x)=x^2, in which case y=x^2-4; clearly not the straight line y=-4!
 
In a graphing calculator, you should graph this cristo: y = f(x) - 4 and you will get a straight line 4 units down. It's going to be a horizontal line and the coordinates would be: (0,-4). I tried it and that's what I got. Maybe I did it wrong but I'm not completely sure about it.
 
Prim3 said:
In a graphing calculator, you should graph this cristo: y = f(x) - 4 and you will get a straight line 4 units down. It's going to be a horizontal line and the coordinates would be: (0,-4). I tried it and that's what I got. Maybe I did it wrong but I'm not completely sure about it.

Your calculator is assuming f(x)=0, since you have not defined it, thus it sees the equation as y=-4 and plots this.

Again, the function f(x) must be specified before it can be sketched.
 
Exactly. But, since he didn't specify what f(x) is, I took it as 0 as well and ended up with y = -4. Wouldn't that work then?
 
  • #10
Prim3 said:
Exactly. But, since he didn't specify what f(x) is, I took it as 0 as well and ended up with y = -4. Wouldn't that work then?

If f(x)=0, then y=f(x)-4 will become y=-4, but why are you taking f(x)=0? Why not take f(x)=17x, or f(x)=x^9? The function f(x) must be specified before the graph can be drawn.
 
  • #11
I'm taking it as a 0 because he didn't specify any other number. If he had specified 1, 2, 3, 4 etc. then I have would've used that. When I was doing this in my homework, an exact question I came upon was y = f(x) - 6 and we had to assume that f(x) = 0 since f(x) wasn't defined.

It's like taking this equation: x^2 + 4x - 5. It's the same as: (1)^2 + 4(1) - 5 because x isn't defined so we take it for 1. Right? At least that's how I learned it.
 
  • #12
Prim3 said:
I'm taking it as a 0 because he didn't specify any other number. If he had specified 1, 2, 3, 4 etc. then I have would've used that. When I was doing this in my homework, an exact question I came upon was y = f(x) - 6 and we had to assume that f(x) = 0 since f(x) wasn't defined.

f(x) isn't a number, it's a function. If you evaluate it at a certain value of x, then it will have a numerical value, but before you do that, it is a function!

Like I said earlier; there is no reason to assume f(x)=0, just because the OP has not told us what it is!

It's like taking this equation: x^2 + 4x - 5. It's the same as: (1)^2 + 4(1) - 5 because x isn't defined so we take it for 1. Right? At least that's how I learned it.

I don't really know what you're doing here. It seems that you are evaluating the function at the point x=1.

We definitely need to know the value of the function before we can sketch the graph (wow, how many times have I said that in this thread? :rolleyes: )
 
  • #13
Yes, I know it's a function. Maybe the way we learned things is different.

Also, for the 2nd part, if you had to say what x was without it being defined at all, what would you say? As far as I know, when it's only a variable (be it x, a, b etc.), we take it as 1.
 
  • #14
Prim3 said:
Also, for the 2nd part, if you had to say what x was without it being defined at all, what would you say?
I'd say it was x. If you're not told the value of x at which to evaluate the function, then you cannot just pick anything!
As far as I know, when it's only a variable (be it x, a, b etc.), we take it as 1.

I'm sure that you're confusing things here. You probably took x=1 when you were required to evalute the function at x=1; i.e. you wanted f(1)
 
  • #15
I was speaking in general and not keeping f(x) in mind regarding the variables being 1. What I'm trying to say is consider this equation: x + 5. If we were to re-write that in another way, it can be 1x + 5. x isn't defined with 1, what I meant to say is explained in that equation. Maybe I said it wrong. Off to bed now; school tomorrow.
 
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  • #16
thanks for your help!
 

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