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
4a2b+5=0
16a+4b+5=59
4a2b+55=05
4a2b=5
16a+4b+55=595
16a+4b=64
5(4a2b)=5*5
20a10b=25
20a10b=25
16a+4b=64
20a10b(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:
4a2b+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 4a2b+5=0, 16a+4b+5=59. Could you please clarify what I need to do next.
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