SOLUTION: Understanding why the graph looks the way it does:
f(x) = (4(x-5))/((x+2)(10-x))
I understand how to find the vertical/horizontal asymptotes and x and y intercept and I under
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-> SOLUTION: Understanding why the graph looks the way it does:
f(x) = (4(x-5))/((x+2)(10-x))
I understand how to find the vertical/horizontal asymptotes and x and y intercept and I under
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Question 916902: Understanding why the graph looks the way it does:
f(x) = (4(x-5))/((x+2)(10-x))
I understand how to find the vertical/horizontal asymptotes and x and y intercept and I understand the middle line going through the middle of the graph but the other two lines curving along the asymptotes why are they there and how do I predict them and know when/where to plot them when I graph on paper? why are they in those positions and why not others?
You can put this solution on YOUR website! Understanding why the graph looks the way it does:
f(x) = (4(x-5))/((x+2)(10-x))
I understand how to find the vertical/horizontal asymptotes and x and y intercept and I understand the middle line going through the middle of the graph but the other two lines curving along the asymptotes why are they there and how do I predict them and know when/where to plot them when I graph on paper? why are they in those positions and why not others?
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Vertical asymptotes at x = -2 and x = 10
Horizontal asymptote at y = x/x^2 = 0/1 = 0
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End Behavior depends on y = x/(-x^2) = -1/x
As x goes to -oo y goes to +0 (above the x-axis)
As x goes to +oo y goes to -0 (below the x-axis)
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Sketching the curve from the left to the right::
Start near zero above the x-axis
Curve up to a VA at x = -2
Signs switch from y positive to y negative at x = -2
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Curve up from the VA asymptote to intersect at (5,0)
Continue up to VA asymptote at x = 10
Signs switch at (x = 10)
Curve up from the VA asymptote to HA (y = 0) below the x-axis.
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Cheers,
Stan H.
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