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# Geometrical applications of Calculus

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The curve y=x3+ ax2+bx+3 has a point of inflexion at (2,-1). Find the values of a and b

Thank you so much

Ashreeta  Dec 4, 2017
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#1
+6250
+1

y  =  x3 + ax2 + bx + 3

We know that when  x = 2  , y''  =  0 .

y'  =  3x2 + 2ax + b

y''  =  6x + 2a                So....

0  =  6(2) + 2a

a  =  -6

Since the graph passes through  (2, -1) , we know...

-1  =  23 + (-6)(2)2 + b(2) + 3

-1  =  8 - 24 + 2b + 3

12  =  2b

b  =  6

Here's a graph:  https://www.desmos.com/calculator/grhajhd2m3

hectictar  Dec 4, 2017
edited by hectictar  Dec 4, 2017
#2
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We need to find the second derivative here

y'  =  3x^2  +  2ax

y" =  6x +  2a       and an inflection point will occur when this = 0

Set this to  0

6x + 2a =  0

3x + a  = 0

3x  =  -a    ⇒    a  = -3x    and when x = 2 we have that  -3(2)  = -6 =  a

So....putting this into the original function to find b, we have that

y  = x^3  + (-6)x^2 + bx  +3

And  we have that at  x = 2, y  = -1....so.....

-1  =  (2)^3 - 6(2)^2 + b(2) + 3

-1 = 8 - 24 + 3 + 2b

-1  = -13 + 2b

12  = 2b   ⇒  b =  6

So the function  is

y  = x^3 -6x^2 + 6x + 3

Here's the graph with the inflection point :

https://www.desmos.com/calculator/6ls40gwylq

CPhill  Dec 4, 2017

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