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

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Answer 1 · 2018-07-28T16:31:44

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

The given function:

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

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

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

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

The above equation shows an upward parabola: ${(x - x_{1})}_{1}^{2} = 4 a (y - y_{1})$ $(x-x_1)^2=4a(y-y_1)$ with

vertex at $(x_{1}, y_{1}) \equiv (1, - 1)$ $(x_1, y_1)\equiv(1, -1)$

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

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

$x (x - 2) = 0$ $x(x-2)=0$

$x = 2, 0$ $x=2, 0$

hence the x-intercept is $x = 2$ $x=2$

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

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

$y = 0$ $y=0$

hence no y-intercept

How do you find the vertex and intercepts for y=x2−2xy=x^2-2x ?