Calculating tangent lines

Function #f# is given by the function rule #f(x)= 3\cdot \euler^{x}+8\cdot x#. Let #k# be the tangent line to the graph of #f# at #x=-4#.

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

#f^{\prime}(x) = 3\cdot \euler^{x}+8 #

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

Give the final formula for #k#

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

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

#3\cdot \euler^ {- 4 }-32 = \left ( 3\cdot \euler^ {- 4 }+8 \right ) \cdot (-4) + b#

#y = \left ( 3\cdot \euler^ {- 4 }+8 \right ) \cdot x + 15\cdot \euler^ {- 4 }#

#f(-4) = 3\cdot \euler^ {- 4 }-32 #

#a = f^{\prime}(-4) = 3\cdot \euler^ {- 4 }+8#

Determine the slope #a# of the tangent

#b = 15\cdot \euler^ {- 4 }#

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