Asked by jayson
What is the equation in standard form of a parabola that models the values in the table?
x -2 0 4
f(x) 1 5 -59
x -2 0 4
f(x) 1 5 -59
Answers
Answered by
Jayson
This is what I have so far if someone could help me finish.
a(-2)^2+b(-2)+c=1
a(0)^2+b(0)+c=5
a(4)^2+b(4)+c=-59
4a-2b+c=1
0a+0b+c=5 c=5
16a+4b+c=-59
4a-2b+5=1
16a+4b+5=-59
4a-2b+5-5=1-5
4a-2b=-4
16a+4b+5-5=-59-5
16a+4b=-64
Now I'm not sure if I'm doing this right.
4(4a-2b)=-4*4
16a+4b=-16
16a-8b=-16
16a+4b=-64
16a-8b-(16a+4b)=-36-(-64)
I'm lost!
Can someone please help?
a(-2)^2+b(-2)+c=1
a(0)^2+b(0)+c=5
a(4)^2+b(4)+c=-59
4a-2b+c=1
0a+0b+c=5 c=5
16a+4b+c=-59
4a-2b+5=1
16a+4b+5=-59
4a-2b+5-5=1-5
4a-2b=-4
16a+4b+5-5=-59-5
16a+4b=-64
Now I'm not sure if I'm doing this right.
4(4a-2b)=-4*4
16a+4b=-16
16a-8b=-16
16a+4b=-64
16a-8b-(16a+4b)=-36-(-64)
I'm lost!
Can someone please help?
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