Which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1?
f(x) = (x – 2i)(x – 3i)
f(x) = (x + 2i)(x + 3i)
f(x) = (x – 2)(x – 3)(x – 2i)(x – 3i)
f(x) = (x + 2i)(x + 3i)(x – 2i)(x – 3i)

f(x) = (x – 2i)(x – 3i)

Step-by-step explanation:

The above funtion gives us roots

x - 2i = 0 , x - 3i = 0

x = 2i , x = 3i

Both the roots have multiplicity 1.

And for leading coefficient, there will be only one variable having power 2 which is x^2 and the coefficient is 1 so the first function fulfills all conditions.

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step-by-step explanation:

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: