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Mathematics 8 Online
OpenStudy (anonymous):

Find the equation of the tangent line to the curve y=(sqrt(x))/(7x+7) at the point (1, f(1)).

Parth (parthkohli):

Find the derivative of the function. The result will be a function that tells you the slope of the tangent at a given \(x\).

Parth (parthkohli):

Do you need help?

OpenStudy (anonymous):

i get 1/14 as the slope. that's with using the x=1 as the f'(x)

OpenStudy (anonymous):

so does the "point" change from (1,f(1)), to (1,1/14)?

Parth (parthkohli):

Hmm, no -- the derivative is the function that tells you the slope of the tangent at any point, so you have to find f'(1). What you got was f(1).

OpenStudy (anonymous):

oh, in that case than I get 0

Parth (parthkohli):

That is correct!

OpenStudy (anonymous):

so the slope is 0

Parth (parthkohli):

Exactly!

OpenStudy (anonymous):

but how do I write a line equation without a slope

OpenStudy (anonymous):

its y=mx+b or y-y1=m(x-x1)

Parth (parthkohli):

Oh, we had to find the equation of the tangent.

Parth (parthkohli):

You know the slope, and a point of the line.

Parth (parthkohli):

That is, (1,0)

OpenStudy (anonymous):

yeah

Parth (parthkohli):

So use y - y1 = m(x - x1)

OpenStudy (anonymous):

y-0=0(x-1)?

Parth (parthkohli):

Yup! Or y = 0.

OpenStudy (anonymous):

so what would the equation of the line be?

Parth (parthkohli):

y = 0.

OpenStudy (anonymous):

my math program is telling me y=0 is incorrect...

Parth (parthkohli):

Hmm... I wonder why.

Parth (parthkohli):

Whoops!

OpenStudy (anonymous):

?

Parth (parthkohli):

The point is not (1,0) but (1,1/14). Sorry.

OpenStudy (anonymous):

so would the equation then look like this? y-1/14=0(x-1)

OpenStudy (anonymous):

then y-1/14=0 then y=1/14

Parth (parthkohli):

Indeed!

OpenStudy (anonymous):

thank you

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