Calculating tangent lines

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Function #f# is given by the function rule #f(x)= 4\cdot \euler^{x}+3\cdot x#. Let #k# be the tangent line to the graph of #f# at #x=-5#.

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

#f^{\prime}(x) = 4\cdot \euler^{x}+3 #

#f(-5) = 4\cdot \euler^ {- 5 }-15 #

Give the final formula for #k#

#b = 24\cdot \euler^ {- 5 }#

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

#y = \left ( 4\cdot \euler^ {- 5 }+3 \right ) \cdot x + 24\cdot \euler^ {- 5 }#

#a = f^{\prime}(-5) = 4\cdot \euler^ {- 5 }+3#

#4\cdot \euler^ {- 5 }-15 = \left ( 4\cdot \euler^ {- 5 }+3 \right ) \cdot (-5) + b#

Determine the slope #a# of the tangent

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

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

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

	Description	Calculation
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6

Differentiation: Applications of derivatives

Calculating tangent lines