Answer
$$\left( {\text{a}} \right)2{x_0} + 3,\,\,\,\,\left( {\text{b}} \right)7$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^2} + 3x + 2,\,\,\,\,{x_0} = 2 \cr
& \left( {\text{a}} \right){\text{Calculating a formula for the slope of the tangent line at }}{x_0} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{\left( {{x_0} + h} \right)}^2} + 3\left( {{x_0} + h} \right) + 2} \right] - \left( {{{\left( {{x_0}} \right)}^2} + 3\left( {{x_0}} \right) + 2} \right)}}{h} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x_0}^2 + 2{x_0}h + {h^2} + 3{x_0} + 3h + 2} \right) - \left( {{x_0}^2 + 3{x_0} + 2} \right)}}{h} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{{x_0}^2 + 2{x_0}h + {h^2} + 3{x_0} + 3h + 2 - {x_0}^2 - 3{x_0} - 2}}{h} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{2{x_0}h + {h^2} + 3h}}{h} \cr
& {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \left( {2{x_0} + h + 3} \right) \cr
& {m_{\tan }} = 2{x_0} + 0 + 3 \cr
& {m_{\tan }} = 2{x_0} + 3 \cr
& \cr
& \left( {\text{b}} \right){\text{Using the formula obtained in part }}\left( {\text{a}} \right){\text{ to find the slope}} \cr
& {\text{at }}{x_0} = 2 \cr
& {m_{\tan }} = 2\left( 2 \right) + 3 \cr
& {m_{\tan }} = 7 \cr} $$
