Find points on the curve y={2x^3+3x^2-12x+1} where the tangent is horizontal?
1 Answer
Oct 10, 2017
(-2,21) (1,6)
Explanation:
step one: find the derivative of the equation.
Step two: Since a horizontal line has a slope of 0, set the derivative to equal 0 and solve.
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.
Step four: write out the coordinates of the points with a slope of zero.
(-2,21) and (1,-6)