What is the tangent equation for 2x3-3x2-8x plus 9 when x equals 2?

The tangent to 2x3 - 3x2 - 8x + 9 at x = 2 is y = 4x - 11

The tangent to y = 2x3 - 3x2 - 8x + 9 at x = 2 has the same gradient as the curve at that point; to find the gradient, differentiate:

dy/dx = 6x2 - 6x - 8

which at x = 2 is:

gradient = 6 x 22 - 6 x 2 - 8 = 4

At x = 2, y = 2 x 23 - 3 x 22 - 8 x 2 + 9 = -3

The equation of a line through point (xo, yo) with gradient m is:

y - yo = m(x - xo)

Thus the equation of the tangent to the line at x = 2 is:

y - -3 = 4(x - 2)

⇒ y = 4x - 11