Answer
$$\eqalign{
& \left( a \right).{\text{ }}m = 9 \cr
& \left( b \right).{\text{ }}y = 9x - 11 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 4{x^2} - 7x + 5;{\text{ }}P\left( {2,7} \right) \cr
& \left( a \right){\text{ Calculate }}f'\left( x \right){\text{ using the definition of the derivative}} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4{{\left( {x + h} \right)}^2} - 7\left( {x + h} \right) + 5 - \left( {4{x^2} - 7x + 5} \right)}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4\left( {{x^2} + 2xh + {h^2}} \right) - 7\left( {x + h} \right) + 5 - 4{x^2} + 7x - 5}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{4{x^2} + 8xh + 4{h^2} - 7x - 7h - 4{x^2} + 7x}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{8xh + 4{h^2} - 7h}}{h} \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left( {8x + 4h - 7} \right) \cr
& f'\left( x \right) = 8x + 4\left( 0 \right) - 7 \cr
& f'\left( x \right) = 8x - 7 \cr
& {\text{Calculate the slope at }}P\left( {2,7} \right) \cr
& m = f'\left( 2 \right) = 8\left( 2 \right) - 7 = 9 \cr
& \cr
& \left( b \right){\text{ Find an equation of the tangent line at }}P\left( {2,7} \right) \cr
& y - 7 = 9\left( {x - 2} \right) \cr
& y - 7 = 9x - 18 \cr
& {\text{ }}y = 9x - 11 \cr} $$
