+0  
 
+1
171
2
avatar+70 

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
 #1
avatar+7096 
+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     smiley

hectictar  Dec 4, 2017
edited by hectictar  Dec 4, 2017
 #2
avatar+86943 
+2

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

 

 

 

cool cool cool

CPhill  Dec 4, 2017

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