How do you find the important parts of the equation to graph the function #y= x^2-x-3#?

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Answer 1 · 2018-07-02T17:01:54

#(-b)/(2a)# gives the axis of symmetry

#(- -1)/(2xx1)#

#1/2#

So #x=1/2# is the line of symmetry

Put #x=1/2# into the equation

#y=(1/2)^2-1/2-3#

#y=-3 1/4#

The vertex is #(1/2, -13/4)#

We know it is a #uu# shaped parabola with a y intercept (0,-3)

Put a couple of values for X into the equation and plot the coordinates then found.