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

2 Answers
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]}

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