Help with Question (1 Viewer)

[Jonneeh]

The curve: y= ax^3 + bx^2 - x + 5
has a point of inflexion at (1,-2).
Find the values of a and b

Thanks

 

[Aerath]

Double dash it.
y' = 3ax^2 + 2bx -1
y'' = 6ax + 2b

There's a point of inflexion when y'' for the x value.
When y'' = 0, x = -b/3a

We know x = 1, hence -b/3a = 1
3a = -b

The point (1,-2) lies on the curve y = ax^3 + bx^2 -x +5
Subbing it in, you get -2 = a + b - 1 + 5
a+b = -6


-2a = -6
a = 3
b = -9

No guarantees - it's been like 2 months since I've done -any- sort of maths.

 

[Jonneeh]

you find y". y' = 3ax^2 +2bx-1....y"= 6ax+2b

if the point (1,-2) has to satisfy this...

-2= 6a(1) +2b (divide by 2 )
-1=3a+b

we need another eqaution to find the values of a and b... since we know that there is a point of inflexion at the point 1,-2... we then sub it back into the main equation and we get...

-2=a(1)^3 + b(1)^2 -1+5

from this we get
a+b = -6....

now solve simultaneously to get a and b =)

hope this helps.


EDIT: beaten to it....^v^ except the point is 1,-2

Ive solved simultaneously but im still not getting the answer...

The Answer in the book says a=3 b=-9

 

[Aerath]

Double dash it.
y' = 3ax^2 + 2bx -1
y'' = 6ax + 2b

There's a point of inflexion when y'' for the x value.
When y'' = 0, x = -b/3a

We know x = 1, hence -b/3a = 1
3a = -b

The point (1,-2) lies on the curve y = ax^3 + bx^2 -x +5
Subbing it in, you get -2 = a + b - 1 + 5
a+b = -6


-2a = -6
a = 3
b = -9

No guarantees - it's been like 2 months since I've done -any- sort of maths.

This. Corrected my answer.

 

[Jonneeh]

This. Corrected my answer.

Thanks =)