with the given roots the equation is p(x) =
a(x-2)^2(x+2)
If the y intercept is at -.8 the you have the point on the graph (0,-.8) and can use this to solve for a
-0.8=a (-2)^2 (2)
-0.8= 8a
a = -0.8/8=-.1
so equate is
p(x) = -.1(x-2)^2((x+2)
Andrew Y.
asked 02/10/21The polynomial of degree 3, P(x), has a root of multiplicity 2 at x=2 and a root of multiplicity 1 at x=−2. The y-intercept is y=−0.8.Find a formula for P(x).
P(x) = ??
with the given roots the equation is p(x) =
a(x-2)^2(x+2)
If the y intercept is at -.8 the you have the point on the graph (0,-.8) and can use this to solve for a
-0.8=a (-2)^2 (2)
-0.8= 8a
a = -0.8/8=-.1
so equate is
p(x) = -.1(x-2)^2((x+2)
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