Calculating tangent lines

Function #f# is given by the function rule #f(x)= 5\cdot \euler^{x}+3\cdot x#. Let #k# be the tangent line to the graph of #f# at #x=2#.

How do we find the formula for #k#? Drag the steps of the solution into the correct order.

#y = \left ( 5\cdot \euler^2+3 \right ) \cdot x -5\cdot \euler^2#

Give the final formula for #k#

#5\cdot \euler^2+6 = \left ( 5\cdot \euler^2+3 \right ) \cdot 2 + b#

Find the #y#-coordinate of the point of tangency

Calculate #b#, the #y#-intercept of the tangent

Determine the slope #a# of the tangent

Enter values of #\rv{x,y}# and #a# in #y=a \cdot x+ b#

#b = -5\cdot \euler^2#

#a = f^{\prime}(2) = 5\cdot \euler^2+3#

#f(2) = 5\cdot \euler^2+6 #

#f^{\prime}(x) = 5\cdot \euler^{x}+3 #

Calculate the derivative #f^{\prime}(x)#