# Show that, for any cubic function of the form y= ax^3+bx^2+cx+d there is a single point of inflection, and the slope of the curve at that point c-(b^2/3a)

*print*Print*list*Cite

### 1 Answer

To find the point of inflection, we need to find the second derivative.

`y=ax^3+bx^2+cx+d =>`

`y'=3ax^2+2bx+c=>`

`y''=6ax+2b`

`y''=0=>6ax+2b=0=>x=(-2b)/(6a)=>x=-b/(3a)`

Since we are discussing cubic function, the assumption is that a is not zero. Thus for any cubic function of the given form, we have the inflection point at `x=-b/(3a)`

To find the slope at this point, we need to plug in the value of x in the first derivative function.

`y'_(x=-b/(3a))=3a(-b/(3a))^2+2b(-b/(3a))+c=>`

`y'=3a*(b^2)/(3a)^2-[2b^2]/(3a)+c=>`

`y'=b^2/(3a)-(2b^2)/(3a)+c=-(b^2)/(3a)+c`

**Thus the slope is** `-(b^2)/(3a)+c`