For the polynomial function ƒ(x) = −x6 + 3x4 + 4x2, find the zeros. then determine the multiplicity at each zero and state whether the graph displays the behavior of a touch or a cross at each intercept.

There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.

You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.

-x^6 + 3x^4 + 4x^2

-x^2(x^4 - 3x^2 - 4)

Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.

-x^2(x^4 - 3x^2 - 4)

-x^2(x^2 - 4)(x^2 + 1)

Now we can use the factors of two perfect squares rule to factor the middle parenthesis.

-x^2(x^2 - 4)(x^2 + 1)

-x^2(x - 2)(x + 2)(x^2 + 1)

We would also want to split the term in the front.

-x^2(x - 2)(x + 2)(x^2 + 1)

(x)(-x)(x - 2)(x + 2)(x^2 + 1)

Now we would set each portion equal to 0 and solve.

First root

x = 0 ---> no work needed

Second root

-x = 0 ---> divide by -1

x = 0

Third root

x - 2 = 0

x = 2

Forth root

x + 2 = 0

x = -2

Fifth and Sixth roots

x^2 + 1 = 0

x^2 = -1

x = +/-

x = +/- i

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.