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November 25, 2015
Posted by **plato** on Tuesday, October 27, 2009 at 12:28am.

f(x)= ((x+5)(x-4)^(2))/((x-2)(x^(4))

- 12th Grade Math -
**Reiny**, Tuesday, October 27, 2009 at 9:05amFirst of all some critical points.

From the single factor of (x+5) at the top, we can say it crosses at (-5,0)

from the double factor of (x-4)^2, the graph will "touch" the x-axis and then reverse direction.

From the x^4 at the bottom, the y-axis will be a vertical asymptote,

From the (x-2)at the bottom, there will be a vertical asymptote at x=2

The highest power at the top will be +x^3 and the highest power at the bottom will be +x^5, so as x approaches infinity in either the positive or negatives, it will approach zero and the graph will approach the x-axis from the top.

I tried some positive and negative values of x close to zero and the function value was negative. Also a value of x = 1.9 gave me a negative value. So the "loop between 0 and 2 lies below the x-axis, suggesting that there is a maximum value for that part of the graph, but still below the x-axis

Also after (4,0), the graph will rise ever so slightly for a maximum just to the right of (4,0) and then approach the x-axis.

The same thing will happen on the left at (-5,). The graph will come up from its y-axis asymptote, cross at (-5,0), rise just ever so slightly and then drop down to approach the x-axis

If you have a graphing calculator, you can zoom in on these critical areas, but on a large scale the small changes near (4,0) and (-5,0) will be hardly noticeable.

- 12th Grade Math -
**stephanie**, Thursday, October 29, 2009 at 10:06pmyou should Pre-Calculus exams grade 12 exams joined week