Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises - Page 151: 53

$$b = 2,\,\,c = 3$$

$$\eqalign{ & {\text{Let }}f\left( x \right) = {x^2} + bx + c \cr & \cr & {\text{Calculate }}f'\left( x \right) \cr & f'\left( x \right) = 2x + b \cr & {\text{The slope at }}x = 1{\text{ is 4, then}} \cr & 4 = 2\left( 1 \right) + b \cr & {\text{Solve for }}b \cr & b = 2 \cr & \cr & {\text{Evaluate the function at }}x = 1, \cr & f\left( 1 \right) = {\left( 1 \right)^2} + b\left( 1 \right) + c \cr & f\left( 1 \right) = 1 + b + c \cr & f\left( 1 \right) = 1 + 2 + c \cr & f\left( 1 \right) = c + 3 \cr & {\text{Point }}\left( {1,c + 3} \right) \cr & \cr & {\text{The equation of the tangent at the point }}\left( {1,c + 3} \right){\text{is}} \cr & y - \left( {c + 3} \right) = 4\left( {x - 1} \right) \cr & y - c - 3 = 4x - 4 \cr & y = 4x - 4 + c + 3 \cr & y = 4x + \left( { - 4 + c + 3} \right) \cr & {\text{Comparing with }}y = 4x + 2 \cr & - 4 + c + 3 = 2 \cr & {\text{Solve for }}c \cr & c = 2 - 3 + 4 \cr & c = 3 \cr} $$