Find the equation of the tangent line to the graph of the given function at the given value of x,
F(x)=((x^2)+28)^(4/5), x=2
$\displaystyle f(x) = (x^2 + 28)^{\frac{4}{5}}$
$\displaystyle f'(x) = \frac{4}{5} 2x (x^2 + 28)^{- \frac{1}{5}}$ by the chain rule
$\displaystyle f'(2) = \frac{16}{5} (32)^{- \frac{1}{5}}$
$\displaystyle f'(2) = \frac{8}{5} $
the rest is pretty routine
$\displaystyle y - f(2) = f'(2) ( x - 2)$
can you finish that off ?