eroshea
find the point where the normal line to y = x + root x at (4,6) crosses the y-axis.
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phi
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did you find the equation for the slope of the curve?
eroshea
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wait.. i'll solve for the slope
anonymous
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take the derivative
replace \(x\) by \(4\)
that gives you the slope
take the negative reciprocal, that give you the slope of the normal line
use the point slope formula with the point \((4,6)\) to find the equation of the normal line \(y=mx+b\)
then \(b\) is where it crosses the y axis
eroshea
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the value of b is the answer then?
anonymous
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yes, that is the y intecept
eroshea
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ok.. thanks.. i'll try solving for it
phi
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Here is a plot of the scenario
eroshea
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thanks!
what i got was 31/5 .. is this correct?
phi
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how ?
eroshea
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i follow the steps satelite gave
phi
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I think you wandered off the path...
eroshea
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why??
phi
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see the plot I posted up above
eroshea
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where is the normal line there?
phi
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the red line
eroshea
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so what do you think is wrong with my answer? is the steps given by satelite somehow have mistakes? i just followed it
phi
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what did you get for the slope ?
phi
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first, what did you get for the derivative?
eroshea
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owh wait.. i got a mistake
eroshea
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b = 46/5 .. i believe that is the correct answer now?
phi
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yes, 9.2 which matches the plot
eroshea
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9.3 :)) thanks so much! help me again.. i'll post a question :)
phi
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9.3 ?
eroshea
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yes.. divide 46 by 5 right ? then it will be 9... remainder is 15.. that will be 3.. so 9.3?
phi
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5*9= 45
eroshea
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lol.. i got a mistake again. sorry.. yes.. 9.2