Asked by plato
How do you graph
f(x)= ((x+5)(x-4)^(2))/((x-2)(x^(4))
f(x)= ((x+5)(x-4)^(2))/((x-2)(x^(4))
Answers
Answered by
Reiny
First 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.
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.
Answered by
stephanie
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