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

2 Answers
Narad T.
Apr 17, 2017

The equation is #y=51x-99#

Explanation:

We calculate the first derivative

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

#f'(x)=6x^2-2x+3#

The equation of the tangent is

#y-f(3)=f'(3)(x-3)#

#f(3)=2*3^3-3^2+3*3=54-9+9=54#

#f'(3)=6*3^2-2*3+3=54-6+3=51#

The tangent is

#y-54=51(x-3)#

#y=51x-99#

graph{(y-2x^3-x^2+2x)(y-51x+99)=0 [-37.5, 35.6, -4.3, 32.23]}

nightowl
Apr 17, 2017

#y = 51x - 99#

Explanation:

When finding the tangent/normal to a line, you always have to find the gradient first.

The equation: #f(x) = 2x^3 - x^2 + 3x# when differentiated will become #f'(x) = 6x^2 - 2x + 3#.

At #x = 3#, the gradient of the curve and the tangent will be 51.

To find the equation of the tangent, you need to use the format #y - y_1 = m ( x - x_1 )# with #m# being your gradient and #x_1 y_1# being your x and y values. We have the x value already, so we need to find the y value by substituting the x value into the original equation.

#2*3^3 - 3^2 + 3*3 = 54#

At (3 , 54) the equation of the tangent would be:
#y - 54 = 51 (x - 3)#

Simplify it by expanding the brackets on the right and the final answer will be
#y = 51x - 99#

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