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?
Well yes, but in the original post he says he doesnt 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.I think the answer kumar wants is:
You take the graph of f(x) and shift it down four units. (Assuming f:R->R).
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!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.
Your calculator is assuming f(x)=0, since you have not defined it, thus it sees the equation as y=-4 and plots this.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.
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.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?
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!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.
I don't really know what you're doing here. It seems that you are evaluating the function at the point x=1.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'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!Also, for the 2nd part, if you had to say what x was without it being defined at all, what would you say?
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)As far as I know, when it's only a variable (be it x, a, b etc.), we take it as 1.