stationary points + points of inflexion (1 Viewer)

[izzy_xo]

hey ,

I was just wondering if anyone could help me with the following question:

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

many thanks.

 

[life92]

hey ,

I was just wondering if anyone could help me with the following question:

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

many thanks.

y = ax^3 + bx^2 - x + 5

y' = 3ax^2 + 2bx - 1

y'' = 6ax + 2b

Since it has an inflexion at (1,-2) then y'' = 0 at (1,-2)

y'' = 6ax + 2b
0 = 6a(1) + 2b
b = -3a

Now sub this and (1,-2) into the original equation and you get...
-2 = a(1)^3 - 3a(1)^2 - 1 + 5
-2 = -2a + 4
-2a = -6
a = 3
Therefore b = -9

 

[izzy_xo]

y = ax^3 + bx^2 - x + 5

y' = 3ax^2 + 2bx - 1

y'' = 6ax + 2b

Since it has an inflexion at (1,-2) then y'' = 0 at (1,-2)

y'' = 6ax + 2b
0 = 6a(1) + 2b
b = -3a

Now sub this and (1,-2) into the original equation and you get...
-2 = a(1)^3 - 3a(1)^2 - 1 + 5
-2 = -2a + 4
-2a = -6
a = 3
Therefore b = -9

Thanks