Find a polynomial function with the zeros -2 multiplicity 2 and 2 multiplicity 2. whose graph passes through the point (-3,50)

Answer 1

So based on your questions the zero multiplicity will set the tone for the generation of the function.

With multiplicity 2 of the zero -2 leads to two terms in the function --> (x+2)²

With multiplicity 2 of the zero 2 leans to two terms in the function --> (x-2)²

The function therefore has a form of (scale and offset)...

y = f(x) = A(x+2)²(x-2)² + B

y = f(x) = A(x²+4x+4)(x²-4x+4) + B

y = f(x) = A(x⁴ - 8x² + 16) + B (after simplification)

The point of interest will lock down the constant or the multiplier (but not both)...

50 = f(3) = A(81 - 72 + 16) + B

50 = f(3) = A(25) + B

Since we have the prerogative of setting one of the two parameters (have only one initial condition)...

Possibility #1:

Let's set B = 0.

50 = f(3) = 25A --> A = 2

f(x) = 2(x⁴ - 8x² + 16)

Possibility #2:

Let's set A = 1.

50 = f(3) = 25 + B --> B = 25

f(x) = (x⁴ - 8x² + 16) + 25 --> (x⁴ - 8x² + 41)

NOTE: With the two parameters there are an infinite number of solutions based on combinations of (A,B) which will satisfy the multiplicity requirements as well as the passing through of a singular point.