## Precalculus (6th Edition) Blitzer

a) $7$ b) $y=7x-8$
a) Now, $m_{\tan}=\lim_\limits{ h\to 0} \dfrac{f(2+h)-f(2)}{h}$ or, $=\lim_\limits{ h\to 0} \dfrac{[2(2+h)^2-(2+h)-(2+h)]-[2(2)^2-2]}{h}$ or, $=\lim_\limits{ h\to 0} \dfrac{2h^2+7h}{h}$ At $(2,6)$; $m_{\tan}=7$ b) The equation of a line is: $y=mx+c$ Now, $y-6=7(x-2) \implies y=7x-8$