points (-2,28) and (1,1) will have a horizontal tangent line.
This problem involves the first derivative of the function and solving it for those cases where the derivative is 0.
So take the expression
y = 2x3 + 3x2 â’ 12x + 8
And for each exponent, multiply the coefficient by the exponent and then subtract 1 from the exponent. Giving.
y = 6x2 + 6x â’ 12
Now you have a quadratic equation, use the quadratic formula to find it's solutions where a = 6, b = 6, and c = -12
The roots are at x = -2 and x = 1
So for the curve y = 2x^3 + 3x^2 â’ 12x + 8 the tangent line is horizontal at x=-2 and x = 1.
The points will be
y = 2(-2)^3 + 3(-2)^2 â’ 12(-2) + 8
y = -16 + 12 + 24 + 8 = 28
(-2,28)
y = 2x^3 + 3x^2 â’ 12x + 8
y = 2 + 3 - 12 + 8 = 1
(1,1)
