Apolynomial factor has a root of -3 with multiplicity 2, a root of 0 with multiplicity 1, a root with of 1 with multiplicity 1, and a root of 3 with multiplicity 2. if the function has a positive leading coefficient and is of even degree, which could be the graph of the function?

Graph A

Step-by-step explanation:

A polynomials is an equation with many terms whose leading term is the highest exponent known as degree. The degree or exponent tells how many roots exist. These roots are the x-intercepts.

This polynomial has roots -3, 0, 1, and 3. This means the graph must touch or cross through the x-axis at these x-values. What determines if it crosses the x-axis or the simple touch it and bounce back? The even or odd multiplicity - how many times the root occurs.

In this polynomial:

Root -3 has even multiplicity of 2 so it only touches and does not cross through.

Root 0 has odd multiplicity of 1 so crosses through.

Root 1 has odd multiplicity of 1 so crosses through.

Root 3 has even multiplicity of 3 so it only touches and does not cross through.

So because of this, only the first and second graphs are possible answers.

Last, what determines the facing of the graph (up or down) is the leading coefficient. If positive, the graph ends point up. If negative, the graph ends point down. All even degree graphs will have this shape.

Only graph 1 matches the features of this polynomial.

`1234567890-=/*-qwertyuiop[]asdfghjkl; '\\zxcvbnm,./+ /

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: