# How do you find the inflection points for g(x)=-x^2+3x+4?

##### 1 Answer
Apr 7, 2015

Your function is a quadratic and is represented, graphically, by a parabola. The only point where your function changes slope or inclination is the vertex, BUT...the concavity does not change! You basically do not have inflection points (The parabola is always a downwards "U").
You can see this by evaluating the second derivative:
$g ' ' \left(x\right) = - 2$ which is constant (no change in concavity) and always negative giving you a downward U (you can also see this considering that the numerical coefficient of ${x}^{2}$ is $- 1$ which is less than zero).

hope it helps,