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

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

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?

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

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### inflexion of a curve in a line

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