Complete the table of values for y = x^2-2x-3

x=3, y=0

x=-1, y=0

Step-by-step explanation:

We will have to factorise to get the value of x before solving for y

Let's solve

y=x^2-2x-3

To factorise x

x^2-3x+x-3

x (x-3) + 1 (x-3)

(x-3) + (x+1)

x-3=0

x=3

x+1=0

x=-1

Let's substitute the value of value to get the value for y

y=x^2-2x-3

When x is 3

y=3^2-2 (3) - 3

y=9-6-3

y=0

When x is-1

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

y=1+2-3

y=0

Therefore when x is 3, y is 0

When x is - 1, y is 0

Anyone know the answer?