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

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Answer 1 · 2018-07-28T16:31:44

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

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