Question

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Answer 1

Vertex form:

y = a(x-h)^2 + k

y = a(x-315)^2 + 630


Another point on this parabola is (0,0). So plug in x = 0 and y = 0 and solve for 'a'


y = a(x-315)^2 + 630

0 = a(0-315)^2 + 630

0 = a(-315)^2 + 630

0 = 99225a + 630

-630 = 99225a

a = -630/99225

a = -2/315

a = -0.0063492063492

a = -0.0063


So

y = a(x-315)^2 + 630

turns into

y = -0.0063(x-315)^2 + 630

y = -0.0063(x^2-630x + 99225) + 630

y = -0.0063x^2 + 3.969x - 625.1175+ 630

y = -0.0063x^2 + 3.969x + 4.88250000000006

Answer 2

You're probably wondering about the + 3.969x + 4.88250000000006 part

well that's from the round off error, if we use a = -0.0063492063492, then we get


y = a(x-315)^2 + 630

y = -0.0063492063492(x-315)^2 + 630

y = -0.0063492063492x^2+4x+6.300000000*10^(-10)

y = -0.0063492063492x^2+4x

which could then become

y = -0.0063x^2+4x

Answer 3

no yours is fine the way it is

Answer 4

lol thx, glad to be of help