# Find the points on the curve y = 2x^{3} + 3x^{2} -12x + 8 where the tangent is horizontal?

**Solution:**

Given

y = 2x^{3} + 3x^{2} - 12x + 8 --->(1)

Differentiate equation (1) w.r.t., x

dy/dx = 6x^{2 }+ 6x - 12 ---> (2)

Equation (2) is ‘slope if tangent at any point’.

Since the tangent is horizontal, its slope is zero.

∴ 6x^{2} + 6x - 12 = 0

x^{2} + x - 2 = 0

(x + 2)(x - 1) = 0

∴ The roots are x = -2 and x = 1

When x = -2,

y = 2(-2)^{3} +3(-2)^{2} -12(-2) + 8

y = -16 + 12 + 24 + 8

y = 28

When x = 1,

y = 2(1)^{3} + 3(1)^{2} -12(1) + 8

y = 2 + 3 - 12 + 8

y = 1

Therefore, required points are (-2, 28) and (1, 1).

## Find the points on the curve y = 2x^{3} + 3x^{2} - 12x + 8 where the tangent is horizontal?

**Summary:**

The points on the curve y = 2x^{3} + 3x^{2} - 12x + 8, where the tangent is horizontal are (-2, 28) and (1, 1).