If #y= x^2 + 2x + 3#, what are the points of inflection, concavity and critical points?

1 Answer
Nov 30, 2016

No point of inflection.
Concavity upwards.
Critical point at #(-1,2)#

Explanation:

Let's calculate the derivatives

#y=x^2+2x+3#

#dy/dx=2x+2#

Critical points, when #dy/dx=0#

#2x+2=0#

#x=-1#

Second derivative

#(d^2y)/dx^2=2#

#(d^2y)/dx^2>0#, so the concavity is upwards

As #(d^2y)/dx^2!=0# there are no points of inflection.

Let's do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-1##color(white)(aaaa)##+oo#

#color(white)(aaaa)##dy/dx##color(white)(aaaaaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##y##color(white)(aaaaaaaaa)##darr##color(white)(aaaa)##uarr#

graph{x^2+2x+3 [-8.49, 7.31, -0.03, 7.87]}