This is how your graph looks like. It DOES tend to positive infinity.
my question gives me a graph that is f'(x) of the function f and asks me to graph f"(x) and a possible f(x)...and i cannot graph f"(x) without knowing the equation of the given graph right? so i was trying to figure out the equation. and it was fairly straight forward because it had roots andwas a polynomial, but at the end (right side) it did not go to positive infinity, but came down towards the x axis...and continued like that off the page..so i assumed it was a horizontal asymptote but if it does not affect the graph much, i could ignore it.
this is the equation i came up with: x(x-2)^2(x-4)(x+2)
but after the last root (4) the graph goes up and thn comes back down and there is a horizontal asyptote at y=0 as x approaches infiinity. how would i show that in the equation?
the graph i have been given, after it rises above the x-axis at x=4, it goes up for some length, then drops back down after x=5,and keeps decreasing, but never equals/passes the x-axis/0...so would that not be a horizontal asymptote? I just want to know how that would be stated within my polynomial equation that i got.
like i think im supposed to use exponential functions (e^x or e^-x) but how would i state that in the equation
Did the graph look anything like the graph in the following link?
I looked around for a function that would work better than e^(-x). e^(-x) will work, but the relative max between x=-2 & x=0 is huge compared the other extrema. So I shifted & flipped the tanh function --- hyperbolic tangent.