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Zokunu

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littlehelp?

Differentiate from first principle y = x^2 +2

=2x

Find the gradient of the secant through the points A (1,3) and B (3,11) on this curve.

=4

Find the equation of the tangent to the curve at the point A

2x - y + 1 =0

Where does the tangent cut the x-axis

x = -1/2

Find the equation of the normal to the curve at the point B.????? <--------This?
 
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What is the relationship between the gradient of the tangent and the gradient of the normal?
 

HeroicPandas

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w8 then what am i doing wrong here?

e) y-11= -1/4 (x-3)

y-11= -1/4x + 3/4

y= -1/4x + 47/4 ?

Answer: x + 6y = 69


u used m = -1/4 (which is the gradient of a line perpendicular to secant AB, which is not the gradient of the normal at B)
 
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Zokunu

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Find the equation of the tangent to the curve at the point A
2x - y + 1 =0

I guess the gradient of the tangent here is 2?

Therefore the gradient of the normal is given by the perpendicular line properties? m1 x m2 = -1?

Is the gradient of the normal -1/2?
 

bedpotato

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Yeah, but the question is asking for the equation of the normal at point B, not A. So find the gradient of the tangent at B (and so on).
 

Zokunu

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Yeah, but the question is asking for the equation of the normal at point B, not A. So find the gradient of the tangent at B (and so on).
mhm, dw guys i got it now :) TY all. I got ONE MORE question related to this so pls help

(a) Differentiate from first principle y = x^3

3x^2

(b) Find the gradient of the secant though the points A and B on this curve, where x=1 and x=3 respectively. ?

Thanks
 

Makematics

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mhm, dw guys i got it now :) TY all. I got ONE MORE question related to this so pls help

(a) Differentiate from first principle y = x^3

3x^2

(b) Find the gradient of the secant though the points A and B on this curve, where x=1 and x=3 respectively. ?

Thanks
You sub in x=1 and x=3 into y=x^3 to get the coordinates A(1,1) and B(3,27)
Then use the gradient formula to get gradient=13
 

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