Norman M. answered 03/06/15
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Lina,
The table wants you to use the values of x which are greater than or equal 0 and less than or equal to 4, so 0, 1, 2, 3, and 4. So you would plug each of these value in place of x in the function.
y=-x2+4x-3
y=-(02)+(4)(0)-3
y=-3
y=-(12)+(4)(1)-3
y=0
y=-(22)+(4)(2)-3
y=1
y=-(32)+(4)(3)-3
y=0
y=-(42)+(4)(4)-3
y=-3
x | y
0 -3
1 0
2 1
3 -2
4 -3
I hope that helps.