Find an equation for the line tangent to the graph ofat the point (2, 0.857142857142857).
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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?
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.