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February 1, 2015

February 1, 2015

Posted by **HELP** on Thursday, September 5, 2013 at 5:11pm.

I know that standard is y=ax^2+bx+c obviously, since the points are on the graph we know x and y so a,b and c need to be found. The vertex is on (4,-4) the points on the left side of the parabola from the vertex up are (3,-3) (2,0) and (1,5). On the right side upwards from the vertex is (5,3) (6,0) and (7,5) I started at the origin to find all the points so if u plot them all and draw te curved lines u will have what I have. Now what do I do???

- Algebra 1 parabola's -
**Reiny**, Thursday, September 5, 2013 at 6:03pmfrom the vertex (4,-4)

we can say that parabola must be

y = a(x-4)^2 -4

to find the value of a, let's use one of the other points

e.g. 1,5)

5 = a(-3)^2 - 4

9 = 9a

a = 1

so the parabola is

y = (x-4)^2 -4 or

y = x^2 - 8x + 12

I will leave it up to you to make sure all those other points we did not use are also on this curve by subbing them into our equation.

They should all satisfy the equation,

e.g. I will use (7,5)

LS = 5

RS = 49 - 56 + 12 = 5 = LS

so far so good!

- Algebra 1 parabola's -
**Taylor**, Thursday, September 5, 2013 at 6:11pmThanks so much!!!

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