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A line of gradient mmm through (a,b)(a,b)(a,b) has point-gradient form [...]\text{[...]}[...].

A

For y=x2−6x+11y=x^2-6x+11y=x2−6x+11, solving dydx=0\frac{dy}{dx}=0dxdy​=0 gives the stationary point's xxx-coordinate as x=3\boxed{x=3}x=3​.

B

A line of gradient mmm through (a,b)(a,b)(a,b) has point-gradient form y−b=m(x−a)\boxed{y-b=m(x-a)}y−b=m(x−a)​.

C

101010

D

The normal to y=x2−6x+11y=x^2-6x+11y=x2−6x+11 at its stationary point is x=3\boxed{x=3}x=3​.

Card 1 of 20

1.9.9 Tangent and normal at a point Flashcards

  1. A Level
  2. /Maths
  3. /1.9.9 Tangent and normal at a point

20 flashcards on OCR (MEI) A Level Maths 1.9.9 Tangent and normal at a point: the key formulae, methods and definitions you need to recall for Component 01, Component 02 and Component 03.

Flashcards