# A(a,b) EQUATIONS OF TANGENTS y = mx +c y = f(x) tangent

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A(a,b) EQUATIONS OF TANGENTS y = mx +c y = f(x) tangent
NB: at A(a, b) gradient of line = gradient of curve gradient of line = m (from y = mx + c ) gradient of curve at (a, b) = f (a) it follows that m = f (a)

Example 24 Find the equation of the tangent to the curve y = x3 - 2x at the point where x = -1. NB: tangent is line so we need point + gradient then we can use the formula y - b = m(x - a) . Point: if x = -1 then y = (-1)3 - (2 X -1) + 1 = (-2) + 1 = 2 point is (-1,2) Gradient: dy/dx = 3x2 - 2 when x = -1 dy/dx = 3 X (-1)2 - 2 = = 1 gradient = 1

Now using y - b = m(x - a) we get y - 2 = 1( x + 1) or y - 2 = x + 1 or y = x + 3

Example 25 Find the equation of the tangent to the curve y = x2 at the point where x = (x  0) Also find where the tangent cuts the X-axis and Y-axis. Point: when x = then y = (-2)2 = 4/4 = 1 point is (-2, 1) Gradient: y = 4x so dy/dx = -8x-3 = x3 when x = -2 then dy/dx = (-2)3 = -8/-8 = 1 gradient = 1

Now using y - b = m(x - a) we get y - 1 = 1( x + 2) or y - 1 = x + 2 or y = x + 3 Axes Tangent cuts Y-axis when x = 0 so y = = 3 at point (0, 3) Tangent cuts X-axis when y = 0 so 0 = x + 3 or x = -3 at point (-3, 0)

Example 26 - (other way round)
Find the point on the curve y = x2 - 6x + 5 where the gradient of the tangent is 14. gradient of tangent = gradient of curve = dy/dx = 2x - 6 so 2x - 6 = 14 or 2x = 20 or x = 10 Put x = 10 into y = x2 - 6x + 5 Point is (10,45) Giving y = = 45

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