Stationary points/Inflexions (1 Viewer)

[dimzi]

Ok i need help with this question....

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

How do you do it? I used the second derivative to give it a go, but it didnt work! Do i have to use simultaneous equations or something?

HELLLP!

 

[sunjet]

2nd derivative

1sec doing it, what are the asnwers so i can check?

 

[shafqat]

remember (1,-2) also lies on the curve
substitute in the function first to get an expression in a and b
do again to the second derivative (which is 0 at the point)
then solve simultaneously

 

[dimzi]

Ummm how do you do that?? I would really appreciate if you wrote it out for me

These last questions of a chapter suck cause they dont give you examples!!

 

[shafqat]

Ok i need help with this question....

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

How do you do it? I used the second derivative to give it a go, but it didnt work! Do i have to use simultaneous equations or something?

HELLLP!

Substitute (1, -2) into curve:
-2 = a + b - 1 + 5
a + b = -6 *

y'' = 6ax + 2b
At (1,-2). y'' = 0 (pt of inflexion)

so 6a + 2b = 0
from * 2a + 2b = -12
subtracting: 4a = 12
a = 3
so b = -9

i think thats right

 

[nor]

Thank You

Yep! thats the answer in the back of the book. The original question was mine, i rang dimzi for help, he posted it on here for me..

thanks shafqat, much appreciated