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Mathematics 17 Online
OpenStudy (zenmo):

How to simplify the point-slope form?

OpenStudy (zenmo):

\[16+\frac{ 1 }{ 6 }(x-1)\]

OpenStudy (misty1212):

lol is this from the last problem ? just asking because that is not right if it is

OpenStudy (zenmo):

yea, i have the answer, but it doesn't count as the solution

OpenStudy (misty1212):

the point was not \((1,8)\) it was \((64,8)\)

OpenStudy (zenmo):

where did u get 8 from?

OpenStudy (misty1212):

is it the last problem? the point was \((64,8)\) you got i t

OpenStudy (zenmo):

sec

OpenStudy (misty1212):

should be \[y-8=\frac{1}{6}(x-64)\]

OpenStudy (misty1212):

maybe i am remembering wrong

OpenStudy (zenmo):

yea but I'm suppose to use linearization form \[f(a)+f'(a)(x-a) \to find the equation\]

OpenStudy (zenmo):

OpenStudy (misty1212):

ok right so \(x_1=64\)

OpenStudy (misty1212):

and \(y_1=f(64)=8\)

OpenStudy (misty1212):

point slope gives \[y-8=\frac{1}{6}(x-64)\]

OpenStudy (zenmo):

isnt 64^2/3 = 16? cube root of 64 is 4, so 4^2=16

OpenStudy (misty1212):

not sure where the 16 or the 1 came from

OpenStudy (misty1212):

ooh sorry

OpenStudy (misty1212):

you are right i am wrong

OpenStudy (misty1212):

but there is still no 1` in it, \(x_1=64\)

OpenStudy (misty1212):

\[y-16=\frac{1}{6}(x-64)\] that's better

OpenStudy (zenmo):

Oh, I found the error, I used 1/6 for a in (x-a) instead of 64. Oops!

OpenStudy (zenmo):

/derp

OpenStudy (misty1212):

ooh

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