# What is the maximum value that the graph of y=-x^2+6x-8?

If the first derivative of the function is $0$ you have a critical point. So we have to solve $y ' = - 2 x + 6 = 0$. This gives $6 = 2 x$ and this gives $x = 3$ as a critical point. Now we can use the second derivative test to check for maximum or minimum. It not always works, but if it does it is easy.
If the second derivative at the critical point is positive, the point is a minimum, if it is negative the point is a maximum. In this case the second derivative is $- 2$, so it is negative everywhere, and the point is a maximum. The function has a maximum in $x = 3$, and the value of the function at $x = 3$ is $1$