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

1 Answer

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

Explanation:

The given function:

y=x^2-2x

y=x^2-2x+1-1

y=(x-1)^2-1

(x-1)^2=(y+1)

The above equation shows an upward parabola: (x-x_1)^2=4a(y-y_1) with

vertex at (x_1, y_1)\equiv(1, -1)

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

0=x^2-2x

x(x-2)=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\cdot 0

y=0

hence no y-intercept