Question

If the graph of y=x^3+ax^2+bx-4 has a point of inflection at (1, -6), find the value of b.

Answer 1

hint: f(x) has a point of inflection when f '' (x) = 0

Answer 2

I know that y"=6x+2a but what's next?

Answer 3

replace y'' with 0, since f '' (x) = 0

Answer 4

replace x with the x coordinate of the point of inflection

Answer 5

So 6x+2a=0? But how do you solve for a?

Answer 6

x = 1 in this case

Answer 7

Because of (1, -6)?

Answer 8

yep, that's the point of inflection

Answer 9

Then what do you do?

Answer 10

6x+2a=0

6(1)+2a=0

6 + 2a = 0

solve for a

Answer 11

Then once you know the value of 'a', you plug it into y=x^3+ax^2+bx-4

Answer 12

Then you plug (1,-6) into y=x^3+ax^2+bx-4 and solve for b

Answer 13

This works because (1,-6) is a point on y=x^3+ax^2+bx-4

Answer 14

Let me try. Don't go away.

Answer 15

alright

Answer 16

Okay, I got the right answer, which is 0. Thanks.

Answer 17

you're welcome