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
SH
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