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

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Presentation on theme: "A(a,b) y = mx +c y = f(x) EQUATIONS OF TANGENTS tangent NB: at A(a, b) gradient of line = gradient of curve gradient of line = m (from y = mx + c ) gradient."— Presentation transcript:

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

2 Example 24 Find the equation of the tangent to the curve y = x 3 - 2x + 1 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 = -1 - (-2) + 1 = 2point is (-1,2) Gradient: dy / dx = 3x when x = -1 dy / dx = 3 X (-1) = = 1 gradient = 1

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

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

5 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 = 0so y = = 3 at point (0, 3) Tangent cuts X-axis when y = 0so 0 = x + 3 or x = -3 at point (-3, 0)

6 Example 26 - (other way round) Find the point on the curve y = x 2 - 6x + 5 where the gradient of the tangent is 14. gradient of tangent = gradient of curve = dy / dx = 2x - 6 so2x - 6 = 14 or 2x = 20 or x = 10 Put x = 10 into y = x 2 - 6x + 5 Givingy = = 45 Point is (10,45)


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