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