Question

Find points on the curve y={2x^3+3x^2-12x+1} where the tangent is horizontal?

Answer 1

(-2,21) (1,6)

Explanation:

step one: find the derivative of the equation.
`y'=6x^2+6x-12`

Step two: Since a horizontal line has a slope of 0, set the derivative to equal 0 and solve.
`y' = 6(x^2+x-2)`
`y' = 6(x+2)(x-1)`
`x= -2, 1`

Step three: plug the x-values found in step 2 back into the original equation to get the y-coordinates of the points on the curve.

`y(-2)= 21`
`y(1)= -6`

Step four: write out the coordinates of the points with a slope of zero.
(-2,21) and (1,-6)