Question

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

Answer 1

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''(x)=-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,