Answer:

From the given equation y=x2−2x−3 , set x=0 then solve for the y-intercept

y=x2−2x−3

y=02−2⋅0−3

y=−3

From the given equation y=x2−2x−3 , set y=0 then solve for the x-intercept

y=x2−2x−3

x2−2x−3=0

by factoring method

(x−3)(x+1)=0

x=3 and x=−1 when y=0

so (3,0) and (−1,0) are x-intercepts

From the given equation y=x2−2x−3 ,by completing the square, find the vertex

y=x2−2x−3

y=x2−2x+1−1−3

y=(x−1)2−4

y−−4