Horizontal inflection point of a parametric polynomial function

[greg_rack]

Homework Statement

Given $y=ax^3+bx^2+2x-3$, find the values of $a$ and $b$ for which the function has an horizontal inflection point at $x=-1$.

Relevant Equations

none

For $x=-1$ to be an *horizontal* inflection point, the first derivative $y'$ in $-1$ must be zero; and this gives the first condition: $a=\frac{2}{3}b$.

Now, I believe I should "use" the second derivative to obtain the second condition to solve the two-variables-system, but how?
Since it is an inflection point, shouldn't even the second derivative be zero? But the fact that it's an horizontal one is confusing me.

 

[etotheipi]

For $x=-1$ to be an *horizontal* inflection point, the first derivative $y'$ in $-1$ must be zero; and this gives the first condition: $a=\frac{2}{3}b$.

Check this

Since it is an inflection point, shouldn't even the second derivative be zero?

Your thinking is correct!

 

[greg_rack]

Your thinking is correct!

But then, why does $y''(-1)=0 \rightarrow b=0$, if the correct result should be $b=2$?

 

[etotheipi]

But then, why does $y''(-1)=0 \rightarrow b=0$, if the correct result should be $b=2$?

I think you made an error in working out the second derivative. Shouldn't it be, $y'(-1) = 3a(-1)^2 + 2b(-1) + 2 = 3a - 2b + 2 \overset{!}{=} 0 \implies 3a = 2b -2$, and then $y''(-1) = 6a(-1) + 2b \overset{!}{=} 0 \implies 3a = b$?

 

[greg_rack]

I think you made an error in working out the second derivative. Shouldn't it be, $y'(-1) = 3a(-1)^2 + 2b(-1) + 2 = 3a - 2b + 2 \overset{!}{=} 0 \implies 3a = 2b -2$, and then $y''(-1) = 6a(-1) + 2b \overset{!}{=} 0 \implies 3a = b$?

Yupp, here's the deal... silly me!
Thanks a lot :)

 

[etotheipi]

Yupp, here's the deal... silly me!
Thanks a lot :)

Don't worry, I do stuff like that all the time