Posted by Nilan on Tuesday, October 4, 2011 at 7:18am.
you are only given two roots, so let the equation be
y = (x+3)(x-5)(x-a)
but you know that (0,1350) lies on it
1350 = (3)(-5)(-a)
1350 = 15a
a = 90
so equation is
y = (x+3)(x-5)(x-90)
expand if you have to, it is more useful in factored form.
btw, your equation of y = ....(x-5)^2 would have a double root of 5, that is, it would "touch" the x-axis at x = 5
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