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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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- Thread starter kumar_23
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In summary, to graph y=f(x)-4, you need to know the original graph of f(x) and then shift it down four units. However, f(x) must be specified before the graph can be drawn. When using a graphing calculator, you can graph y=f(x)-4 by inputting the equation and shifting it down four units, but the function f(x) must be defined beforehand. It is not correct to assume f(x)=0 or any other value without being specified.

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

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ZioX said:

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:

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!

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Prim3 said:

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.

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Prim3 said:

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.

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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.

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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 toassumethat 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? )

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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.

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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!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?

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)

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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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thanks for your help!

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