Question

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

Answer 1

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]}

Answer 2

`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`