# How do you find the vertex and the intercepts for y=x^2-2x+6?

Vertex is at $\left(1 , 5\right)$ with y-intercept is at $\left(0 , 6\right)$ and having no x-intercept.
$y = {x}^{2} - 2 x + 6 \mathmr{and} y = {\left(x - 1\right)}^{2} \pm 1 + 6 \mathmr{and} y = {\left(x - 1\right)}^{2} + 5$ Comparing with standard equation of parabola $y = a {\left(x - h\right)}^{2} + k$ we find vertex as $\left(h , k\right) \mathmr{and} 1 , 5$
To find y intercept putting $x = 0$ in the equation we get $y = 6$
To find x intercept putting $y = 0$ in the equation we get ${x}^{2} - 2 x + 6 = 0 \mathmr{and} {\left(x - 1\right)}^{2} = - 5 \mathmr{and} \left(x - 1\right) = \pm \sqrt{- 5} \mathmr{and} x = 1 \pm \sqrt{5} i \therefore x$ has complex roots and there will be no x-intercept. graph{x^2-2x+6 [-45, 45, -22.5, 22.5]}[Ans]