The curve above is the graph of a degree 4 polynomial. It goes through the point (5, -175). First determine the intercepts and factors, then the lead coefficient 'a' and finally enter the polynomial F(x)=___?

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Oddman answered
The zeros appear to be {-2, 0, 0, +3}. These x-intercepts suggest the factors   F'(x) = (x+2)(x^2)(x-3) When we evaluate F'(5), we get
  F'(5) = (7)(25)(2) = 350
We know F(5) = -175, so there must be an additional factor (lead coefficient 'a') of -175/350 = -1/2.
The polynomial is
  F(x) = -1/2(x+2)(x^2)(x-3)
  F(x) = -1/2x^4 + 1/2x^3 + 3x^2

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