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

Vertex $\left(1 , - 1\right)$, x-intercept $x = 2$ & no y-intercept

#### Explanation:

The given function:

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

$y = {x}^{2} - 2 x + 1 - 1$

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

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

The above equation shows an upward parabola: ${\left(x - {x}_{1}\right)}^{2} = 4 a \left(y - {y}_{1}\right)$ with

vertex at $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(1 , - 1\right)$

x-intercept: setting $y = 0$ in given equation to get x-intercept as follows

$0 = {x}^{2} - 2 x$

$x \left(x - 2\right) = 0$

$x = 2 , 0$

hence the x-intercept is $x = 2$

y-intercept: setting $x = 0$ in given equation to get y-intercept as follows

$y = {0}^{2} - 2 \setminus \cdot 0$

$y = 0$

hence no y-intercept