Graphing Question: Finding Roots and Horizontal Asymptote of x(x-2)^2(x-4)(x+2)

[menal]

Homework Statement

I have a graph and its roots are:
x(x-2)^2(x-4)(x+2)

^that is also the equation i got after viewing the graph. 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?

 

[HallsofIvy]

Unless you have miswritten the function, there are NO "asymptotes"- that is a polynomial.

What I would do is rearrange the factors so each term, in the form x- a, is increasing a. That is, write it as [itex](x+ 2)x(x- 2)^2(x- 4)[/tex] (x+ 2= x- (-2) and x= x- 0).

That is a polynomial that crosses the x-axis at x= -2, x= 0, x= 2, and x= 4. To see what it looks like between those points, remember that if x< a, then x-a< 0 but if x> a then x-a> 0. Start, on the left, with x< -2. The all of those 5 factors (we are counting x-2 twice, of course) is negative and the product of an odd number (5) of negative numbers is negative. For x< -2, this function is negative. Its graph rises from $-\infty$ on the left to (-2, 0).

If -2< x< 0, then x+2> 0 but all other four factors are negatve. The product of 4 negative factors is positive and it product with a positive factor is postive. The graph rises above the x-axis to some maximum, then back to (0, 0).

If 0< x< 2, then both x+ 2> 0 and x> 0. We now have two positive factors and three negative factors which is negative. The graph drops below the x-axis at (0, 0), down to some minimum, then back up to (2, 0).

If 2< x< 4, then we have four positive factors (they jumped from two to four because the x-2 factor is squared) and one negative factor. The product remains negative. So the graph does NOT cross the x-axis at (2, 0). The graph "kisses" it (is tangent to the x-axis) then drops back down to some negative minimum then back up to (4, 0).

Finally, if x> 4, all five factors are positive but the product of any number of positive factors is positive. The graph rises above the x-axis at x= 4, going up, on the right, toward positive infinity. (The graph does NOT come back down, there is NO horizontal asymptote!)

 

[menal]

yes, that is all correct, except 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.

 

[SammyS]

What does it look like along the negative x-axis?

 

[menal]

it goes toward negative infinity, then does all the things you stated, and ends with the asymptote that i mentioned.
so basically it looks like a normal polynomial until x=5.

 

[SammyS]

What function do you know which goes to -∞ as x goes to -∞, and goes to zero as x goes to +∞ . Also, it has no x intercepts.

 

[menal]

rational functions?

 

[SammyS]

No. A rational function would have the same horizontal asymptote on the left & right, or else no horizontal asymptote.

 

[menal]

i have no clue

 

[SammyS]

How about an exponential or logarithmic function?

 

[menal]

i know it cannot be logrthmic because we have to find the second derivative...and we havnt learned how to find the derivative of logs...(if there is one). so its exponential, but we already have exponents in the equation already..

 

[SammyS]

An exponential function has x as the exponent.

What you have is a polynomial. Polynomials have natural numbers as the exponents, with x as the base.

Two basic forms of the exponential function are: e^x and e^-x.