# How do you find the axis of symmetry, and the maximum or minimum value of the function y = x^2 - 4x + 1?

The axis of symmetry is $x = 2$ and minimum point is at $\left(2 , - 3\right)$.
$y = {x}^{2} - 4 x + 1 \mathmr{and} y = {\left(x - 2\right)}^{2} - 4 + 1 \mathmr{and} y = {\left(x - 2\right)}^{2} - 3$.The vertex is at $\left(2 , - 3\right)$.As the sign of ${x}^{2}$ is +ive, the parabola opens upwards.
The axis of symmetry is $x = 2$ and minimum point is at $\left(2 , - 3\right)$. graph{x^2-4x+1 [-10, 10, -5, 5]}[Ans]