Equations of Tangents.

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Presentation transcript:

Equations of Tangents

What is to be learned How to use differentiation to find all the stuff we need to get equation of a tangent.

tangent at x = 3? y = x2 For Equation need gradient m point (a , b) 3

tangent at x = 3? y = x2 dy/dx = 2x at x = 3 m = 2 X 3 so m = 6 For gradient find The Derivative 3

tangent at x = 3? y = x2 (3 , ?) Point? Need y coordinate 3

tangent at x = 3? y = x2 at x = 3 y = 32 (3 , ?) so (a , b) = (3 , 9) Point? Need y coordinate 3

tangent at x = 3? y = x2 (3 , ?) m = 6, (a , b) = (3 ,9) Need y coordinate 3 use y - b = m(x – a)

Usually not given diagram! Ex For curve y = x3 – 2x2 + 5 Find equation of tangent at x = 3 Get m dy/dx = 3x2 – 4x so at x = 3 m = 3(3)2 – 4(3) = 15 → Need Derivative

Usually not given diagram! Ex For curve y = x3 – 2x2 + 5 Find equation of tangent at x = 3 Get point so at x = 3 y = 33 – 2(3)2 + 5 = 14 (a , b) = (3 , 14) Then y – b = m(x – a) → Use equation, y =

Equations of Tangents For Equation need gradient m point (a , b) Find Derivative Use original equation Then use y - b = m(x – a)

tangent at x = 4? y = x2 – 6x 4

tangent at x = 4? y = x2 – 6x dy/dx = 2x - 6 at x = 4 m = 2(4) - 6 so m = 2 For gradient find The Derivative 4

tangent at x = 4? y = x2 – 6x Point? 4 Need y coordinate (4 , ?)

tangent at x = 4? y = x2 – 6x at x = 4 y = 42 – 6(4) = -8 so (a , b) = (4 , -8) Point? 4 Need y coordinate (4 , ?)

tangent at x = 4? y = x2 – 6x m = 2, (a , b) = (4 , -8) 4 use y - b = m(x – a)