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The gradient, 𝑑𝑦/𝑑𝑥 , of a curve C at the point (𝑥, 𝑦) is given by 𝑑𝑦/𝑑𝑥 = 20𝑥 − 6𝑥^2 − 16 The curve C passes through the point P (2, 3).

a) Verify that the tangent to the curve at P is parallel to the x-axis.

b) The point Q(3, -1) also lies on the curve. The Normal to the curve at Q and the tangent to the curve at P intersect at the point R. Find the coordinates of R

 Dec 3, 2018
 #1
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a)   f ' (2) =  20(2) - 6(2)^2 - 16  =  40 - 24 - 16 = 0

= the slope of the x axis......so.....parallel

 

The equation of this tangent line is y = 0 (x -2) + 3 .... y = 3         (1)

 

b)  The slope of the tangent line at x = 3 is

20(3) - 6(3)^2 - 16 =  60 - 54 - 16 =  -10

So....the slope of the normal line is (1/10)

The equation of this normal line is

y = (1/10)(x - 3) -1

y = (1/10)x - 3/10 - 1

y = (1/10)x - 13/10          sub (1) into this to find the x coordinate of R

 

3 = (1/10)x - 13/10

3 + 13/10 = x /10

43/10 = x /10

43 = x

 

So.....R = (43, 3)

 

See the graph here to confirm this :

 

https://www.desmos.com/calculator/0ipyvcpc5y

 

P.S. - If you have had Integral Calculus....you might see how I determined f(x)

 

 

cool cool cool

 Dec 3, 2018

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