What is the equation of the tangent line of #f(x)=x^3+2x^2-3x+2# at #x=1#?

1 Answer
Dec 24, 2015

#y = 4x -2#

Explanation:

Step 1: Find derivative of the equation

#f(x) = x^3 + 2x^2 -3x +2#

#f'(x) = d/dx(x^3) + d/dx(2x^2) -d/dx(3x) + d/dx(2) #

#f'(x) = 3x^2 + 4x -3#

Step 2: Find the slope of the tangent line at #x = 1#
#m = f'(1) = 3(1^2) +4(1) -3 = 4#

Step 3: Find the #y# coordinate of the function when #x= 1#

#f(1) = (1)^3 + 2(1)^2 -3(1) +2#

#f(1) = 2#

So, the original point of the graph is #(1,2)#

Step 4: Find the equation of the tangent line using point slope formula

#y -y_0 = m(x-x_0)#

# m= 4 " " " " (1, 2)#

#y-2 = 4(x - 1)#
#=> y-2 = 4x - 4#
#=> y = 4x -4 +2#
#=> y = 4x -2#