Posted by **Lee** on Monday, October 29, 2012 at 4:28pm.

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

Please help.I have somewhat of an idea

a(-2)^2+b(-2)+c c=1

a(0)^2+b(0)+c= c=5

a(4)^2+b(4)+c c=-59

4a+2b+c=1

0a+0b+c=5

16a+4b+c=-59

4a-2b+5=0

16a+4b+5=-59

4a-2b+5-5=0-5

4a-2b=-5

16a+4b+5-5=-59-5

16a+4b=-64

5(4a-2b)=5*5

20a-10b=25

20a-10b=25

16a+4b=-64

20a-10b-(16a+4b)=25-(64)

4a+14b=89

not sure what to do now

- algebra -
**Steve**, Monday, October 29, 2012 at 4:41pm
after subbing in c=5, you got here:

4a-2b+5=0

16a+4b+5=-59

You made a couple of errors here. You should have come up with:

4a - 2b = -4

16a + 4b = -64

or,

4a - 2b = -4

4a + b = -16

b = -4

a = -3

So, y = -3x^2 - 4x + 5

Take much care when proceeding from step to step, that you don't make transcription errors!

- algebra -
**Lee**, Monday, October 29, 2012 at 9:47pm
Not sure where I made my mistake.I lost myself after the 4a-2b+5=0, 16a+4b+5=-59. Could you please clarify what I need to do next.

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