# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function y = -x^2 + 2x?

Feb 28, 2018

$\left(1 , 1\right)$ $\to$ local maximum.

#### Explanation:

Putting the equation in vertex form,

$y = - {x}^{2} + 2 x$

$y = - \left[{x}^{2} - 2 x\right]$

$y = - \left[{\left(x - 1\right)}^{2} - 1\right]$

$y = - {\left(x - 1\right)}^{2} + 1$

In vertex form, the $x$ coordinate of the vertex is the value of $x$ which makes the square equal to $0$, in this case, 1 (since ${\left(1 - 1\right)}^{2} = 0$).

Plugging this value in, the $y$ value turns out to be $1$.

Finally, since it is a negative quadratic, this point $\left(1 , 1\right)$ is a local maximum.