Find the tangent to y = x^2 - 4x + 9 at the point (3,15)

First find dy/dx:

dy/dx = 4x - 4

And thus at (3,15):

dy/dx = 12 - 4 = 8 = m (as m is the gradient of a curve)

So using y - y₁ = m(x - x₁) where (x₁,y₁) = (3,15):

y - 15 = 8(x - 3)

y = 8x- 9