Answer
a) $\dfrac{1}{6}$
b) $ y=\dfrac{1}{6}x+\dfrac{3}{2}$
Work Step by Step
a) Now, $ m_{\tan}=\lim_\limits{ h\to 0} \dfrac{f(9+h)-f(9)}{h}$
or, $=\lim_\limits{ h\to 0} \dfrac{\sqrt{9+h}-\sqrt 9}{h}$
or, $=\lim_\limits{ h\to 0} \dfrac{1}{\sqrt {9+h}+3}$
At $(9,3)$; $ m_{\tan}=\dfrac{1}{6}$
b) The equation of a line is: $ y=mx+c $
Now, $ y-3=(1/6)(x-9) \implies y=\dfrac{1}{6}x+\dfrac{3}{2}$
