Question

How do you find the equation of the tangent line to the curve `y=3x^2-x^3` at point (1,2)?

Answer 1

`y=3x-1`

Explanation:

Find `dy/dx` of `y` equation

`y=3x^2-x^3`
`dy/dx=6x-3x^2`
Then from the point (1,2) we can obtain the x value which is 1
Insert `x=1` into `dy/dx`
`dy/dx=6(1)-3(1)^2`
`dy/dx=3`

To form the equation we have to base it from the original equation
`y=mx+c`
we know that m=3 from `dy/dx=3`
`y=3x+c`
use the coordinates given fro x and y values to find c
(1,2) x=1 , y=2
`(2)=3(1)+c`
`c=-1`
Then we put c back into the general form to get the final tangent equation
`y=3x-1`