Question

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

Answer 1

y = 4x - 2

Explanation:

To find the equation of the tangent require gradient ( m ) and a
point on the line (a , b ).

The derivative of y `( dy/dx ) = m`

To find a point on the line use x = 1 substituted into equation
to find y coordinate.

`dy/dx = 3x^2 + 4x - 3`

when x = 1 : `dy/dx = 3(1)^2 + 4(1) - 3 = 3 + 4 - 3 = 4`

and `y = (1)^3 + 2(1)^2 - 3(1) + 2 = 1 + 2 - 3 + 2 = 2`

equation if tangent is : y - b =m(x - a ) where m = 4 and

( a , b ) = (1 , 2 )

hence y - 2 = 4( x - 1 ) → y - 2 = 4x - 4

`rArr y = 4x - 2`