1. Deliveries 3 2 2 X 4 6 Measurement
2. Deliveries 3 2 2 X 4 Equal
3. 5 X 4 3 X 2
4. Deliveries 3 2 2 X 4

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

1 Answer

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#

Deliveries 3 2 2 X 4 6 Measurement

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# Affinity designer 1 5 1 – vector graphic design software.

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

Step 4: Find the equation of the tangent line using point slope formula Elements for iwork 3 1 download free.

Deliveries 3 2 2 X 4 Equal

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

5 X 4 3 X 2

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

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

Deliveries 3 2 2 X 4

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