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

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Question:

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

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For the values of 'x' that is -3, -2, -1, 0, 1, 2, 3, 4, 5 the values of 'y' is 12, 5, 0, -3, -4, -3, 0, 5, 12 respectively and this can be determined by factorising the given quadratic equation and then put the values of 'x' in it.Given :Quadratic equation - [tex]y = x^2-2x-3[/tex]To complete the table, first, factorize the given quadratic equation.[tex]y = x^2-2x-3[/tex][tex]y = x^2-3x+x-3[/tex][tex]y = (x-3)(x+1)[/tex]     ----- (1)Now, according to the given table the values of 'x' is:x = -3, -2, -1, 0, 1, 2, 3, 4, 5Put the value of x=-3 in equation (1).y = (-3-3)(-3+1)y = 12Put the value of x=-2 in equation (1).y = (-2-3)(-2+1)y = 5Put the value of x=-1 in equation (1).y = (-1-3)(-1+1)y = 0Put the value of x=0 in equation (1).y = (0-3)(0+1)y = -3Put the value of x=1 in equation (1).y = (1-3)(1+1)y = -4Put the value of x=2 in equation (1).y = (2-3)(2+1)y = -3Put the value of x=3 in equation (1).y = (3-3)(3+1)y = 0Put the value of x=4 in equation (1).y = (4-3)(4+1)y = 5Put the value of x=5 in equation (1).y = (5-3)(5+1)y = 12For more information, refer to the link given below:https://brainly.com/question/13911928

Answer:x=3,y=0x=-1,y=0Step-by-step explanation:We will have to factorise to get the value of x before solving for yLet's solvey=x^2-2x-3To factorise xx^2-3x+x-3x(x-3)+1(x-3)(x-3)+(x+1)x-3=0x=3x+1=0x=-1Let's substitute the value of value to get the value for yy=x^2-2x-3When x is 3y=3^2-2(3)-3y=9-6-3y=0When x is-1y=(-1)^2-2(-1)-3y=1+2-3y=0Therefore when x is 3,y is 0When x is -1,y is 0

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