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
Explanation:
Step 1: Find derivative of the equation
#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
Step 3: Find the
#f(1) = 2#
So, the original point of the graph is
Step 4: Find the equation of the tangent line using point slope formula
#y -y_0 = m(x-x_0)#
# m= 4 " " " " (1, 2)#