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Higher Maths - Equation Of A Tangent

05/09/2022

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16.8.21
dy
dx
Equation of a langent./
M
-= dy
de
ہے
tangent
Therefore
we need to know two things about the tangent.
given.
point, of which a

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16.8.21
dy
dx
Equation of a langent./
M
-= dy
de
ہے
tangent
Therefore
we need to know two things about the tangent.
given.
point, of which a

Register

Sign up to get unlimited access to thousands of study materials. It's free!

Access to all documents

Join milions of students

Improve your grades

By signing up you accept Terms of Service and Privacy Policy

16.8.21
dy
dx
Equation of a langent./
M
-= dy
de
ہے
tangent
Therefore
we need to know two things about the tangent.
given.
point, of which a

Register

Sign up to get unlimited access to thousands of study materials. It's free!

Access to all documents

Join milions of students

Improve your grades

By signing up you accept Terms of Service and Privacy Policy

16.8.21 dy dx Equation of a langent./ M -= dy de ہے tangent Therefore we need to know two things about the tangent. given. point, of which at least one coordinate will be • the gradient, which is calculated by differentiating and substitulay x at the required point. in the value of Examples; 1. Find the equation of the langent to the curve with equation y= x²-3 at the point (2.1). ( dy = 2x de To work out the equation of a tangent y-b= m(x-a). m = 2x we use M = 2x2 m = 4 y-b= a (x-2) y-1 = 4(x-2) y-1 = 4x=8 y = 4x-7 2. Find the equation of the langent to the y = x² - 2x at the point where x=-1. Find y X1-2x y = y = (¹1) ³ - 2 (-¹1) y = -1 +2 y=1 (-11) 19.8.21 m= dy de Y- 8 = 11 (x - 1) 4- 8 = 11 x - 11 y=11x - 3 Gradient. dy dx dy = ME Equation of a targent.... cont. dy dx M = 3(-1) ²-1 m = 3-2 M = 1 6x + 5 3x²-2 Find the equation of the langeet to the curve gradient of the Langent is 11. 6x + 5 = 11 6x = 6 x=1 curve wife wild Equation: y=1=1 (x + 1) y-1 = x+1 y=x+2 y = 3x² + 5x ²/1 y = 3 (1) ² + 5 (1) y = 8 (1.8) Note: Find gradi I tangent x² = More Langents x² = For y U 23 M Find the equation of the tangents to the curve y=x²-4x when the gradient of the langents dy dx For x = -2 y = -2² -4 (-2) y = 0 (-2,0) dy dx =...

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Alternative transcript:

3x² - 4 3x²-4=8 3x² = 12 X² = 4 x = √√4= .8 رنا OR: 3x²-12=0 3 (x² - 4) = 0 3 (x-2)(x+2) = 0 x = 2 x = -2 y-0 = 8(x - 2) y=8x-16 For x=2 Y = 2³ - 4 (2) y=0 (2,0) y-0. = 8(x+2) y = 8x + 16