equation of the line tangent to a curve at its inflection point

**LexiRae** · February 2nd 2009, 11:55 AM

An equation of the line tangent to the graph of y=a^3+ax^2+2 at its point of inflection is:?

i have no idea.
i know you do something with the second derivative and find where it goes from concave upward to concave downward.
but whats with the equation?
thanks in advance!

**Plato** · February 2nd 2009, 02:09 PM

Originally Posted by LexiRae

An equation of the line tangent to the graph of y=a^3+ax^2+2 at its point of inflection is:?

I am sure that you mean $a {\color{red}x^3} +ax^2 +2$
Slope: $m=3ax^2+2ax$ .
Second derivative: $6ax+2a$
Then inflection point: $6x + 2a = 0\; \Rightarrow \;x = \frac{{ - 1}}{3}$ .
The point is $\left( {\frac{{ - 1}}{3},\frac{{2a}}{{27}} + 2} \right)$ and slope is $m=\frac{-a}{3}$
You can finish.

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