Find an equation for the line tangent to the graph ofhttps://webwork.math.lsu.edu/webwork...1dc2bdbc81.png at the point (2, 0.857142857142857).

https://webwork.math.lsu.edu/webwork...22b824af51.png =

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- Oct 3rd 2008, 01:30 PMjohnny4lsuTangen Line Equation Problem???
Find an equation for the line tangent to the graph of

https://webwork.math.lsu.edu/webwork...1dc2bdbc81.png at the point (2, 0.857142857142857).

https://webwork.math.lsu.edu/webwork...22b824af51.png = - Oct 3rd 2008, 01:36 PMskeeter
see this thread ...

http://www.mathhelpforum.com/math-he...gent-line.html

btw, why the decimal for $\displaystyle \frac{6}{7}$ ?

you uncomfortable with fractions? - Oct 3rd 2008, 01:38 PMjohnny4lsu
this is what they gave me in the problem

- Oct 3rd 2008, 01:43 PMjohnny4lsu
I tried to use that link but i am still coming up with the wrong answer. (Headbang)

could someone possible walk me through it? - Oct 3rd 2008, 01:51 PMskeeter
what did you get for the derivative of $\displaystyle f(x)$?

- Oct 3rd 2008, 01:54 PMjohnny4lsu
15/(x+5)^2

- Oct 3rd 2008, 02:04 PMskeeter
good.

the slope of the tangent line is $\displaystyle f'(2) = \frac{15}{(2+5)^2} = \frac{15}{49}$

use the point-slope form for a linear equation ...

$\displaystyle y - \frac{6}{7} = \frac{15}{49}(x - 2)$

this is the tangent line equation in point-slope form.

if you need to change it to standard form, multiply both sides by 49 to clear the fractions and manipulate accordingly. - Oct 3rd 2008, 02:10 PMjohnny4lsu
Thanks again skeeter. I was so close!!!