QuestionDetermine the end behavior of $f(x)=-5 x^{2}(x^{2}-11)$ and find its real zeros and their multiplicities.

Studdy Solution

STEP 1

Assumptions1. The function is given by $f(x)=-5x^{}(x^{}-11)$ . We are asked to find the real zeros of the function, their multiplicities, the behavior of the graph at the x-intercepts, and the maximum number of turning points.

STEP 2

To find the real zeros of the function, we need to set the function equal to zero and solve for $x$ .
$-5x^{2}(x^{2}-11) =0$

STEP 3

We can solve this equation by setting each factor equal to zero.
$-5x^{2} =0 \quad \text{and} \quad x^{2}-11 =0$

STEP 4

olving the first equation gives us one of the real zeros.
$-x^{2} =0 \implies x =0$

STEP 5

olving the second equation gives us the other two real zeros.
$x^{2}-11 =0 \implies x = \sqrt{11}, -\sqrt{11}$

STEP 6

The multiplicities of the zeros are the powers of the corresponding factors in the factored form of the polynomial. The factor $x^{2}$ corresponds to the zero $x =0$ , and it has a power of2, so the multiplicity of this zero is2.

STEP 7

The factors $x - \sqrt{11}$ and $x + \sqrt{11}$ correspond to the zeros $x = \sqrt{11}$ and $x = -\sqrt{11}$ , respectively, and they each have a power of1, so the multiplicities of these zeros are1.

STEP 8

The graph of a polynomial function crosses the x-axis at a zero if the zero has an odd multiplicity, and touches the x-axis and turns around at a zero if the zero has an even multiplicity.

STEP 9

So, the graph of $f$ crosses the x-axis at the zeros $x = \sqrt{11}$ and $x = -\sqrt{11}$ , and touches the x-axis and turns around at the zero $x =$ .

STEP 10

The maximum number of turning points of a polynomial function is one less than the degree of the function. The degree of $f(x)=-5x^{2}(x^{2}-)$ is4, so the maximum number of turning points is4 - =3.

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QuestionDetermine the end behavior of $f(x)=-5 x^{2}(x^{2}-11)$ and find its real zeros and their multiplicities.

Studdy Solution

STEP 1

Assumptions1. The function is given by $f(x)=-5x^{}(x^{}-11)$ . We are asked to find the real zeros of the function, their multiplicities, the behavior of the graph at the x-intercepts, and the maximum number of turning points.

STEP 2

To find the real zeros of the function, we need to set the function equal to zero and solve for $x$ .
$-5x^{2}(x^{2}-11) =0$

STEP 3

We can solve this equation by setting each factor equal to zero.
$-5x^{2} =0 \quad \text{and} \quad x^{2}-11 =0$

STEP 4

olving the first equation gives us one of the real zeros.
$-x^{2} =0 \implies x =0$

STEP 5

olving the second equation gives us the other two real zeros.
$x^{2}-11 =0 \implies x = \sqrt{11}, -\sqrt{11}$

STEP 6

The multiplicities of the zeros are the powers of the corresponding factors in the factored form of the polynomial. The factor $x^{2}$ corresponds to the zero $x =0$ , and it has a power of2, so the multiplicity of this zero is2.

STEP 7

The factors $x - \sqrt{11}$ and $x + \sqrt{11}$ correspond to the zeros $x = \sqrt{11}$ and $x = -\sqrt{11}$ , respectively, and they each have a power of1, so the multiplicities of these zeros are1.

STEP 8

The graph of a polynomial function crosses the x-axis at a zero if the zero has an odd multiplicity, and touches the x-axis and turns around at a zero if the zero has an even multiplicity.

STEP 9

So, the graph of $f$ crosses the x-axis at the zeros $x = \sqrt{11}$ and $x = -\sqrt{11}$ , and touches the x-axis and turns around at the zero $x =$ .

STEP 10

The maximum number of turning points of a polynomial function is one less than the degree of the function. The degree of $f(x)=-5x^{2}(x^{2}-)$ is4, so the maximum number of turning points is4 - =3.

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QuestionDetermine the end behavior of f(x)=−5x2(x2−11)f(x)=-5 x^{2}(x^{2}-11)f(x)=−5x2(x2−11) and find its real zeros and their multiplicities.

Studdy Solution

STEP 1

Assumptions1. The function is given by f(x)=−5x(x−11)f(x)=-5x^{}(x^{}-11)f(x)=−5x(x−11). We are asked to find the real zeros of the function, their multiplicities, the behavior of the graph at the x-intercepts, and the maximum number of turning points.

STEP 2

To find the real zeros of the function, we need to set the function equal to zero and solve for xxx.−5x2(x2−11)=0-5x^{2}(x^{2}-11) =0−5x2(x2−11)=0

STEP 3

We can solve this equation by setting each factor equal to zero.−5x2=0andx2−11=0-5x^{2} =0 \quad \text{and} \quad x^{2}-11 =0−5x2=0andx2−11=0

STEP 4

olving the first equation gives us one of the real zeros.−x2=0 ⟹ x=0-x^{2} =0 \implies x =0−x2=0⟹x=0

STEP 5

olving the second equation gives us the other two real zeros.x2−11=0 ⟹ x=11,−11x^{2}-11 =0 \implies x = \sqrt{11}, -\sqrt{11}x2−11=0⟹x=11​,−11​

STEP 6

The multiplicities of the zeros are the powers of the corresponding factors in the factored form of the polynomial. The factor x2x^{2}x2 corresponds to the zero x=0x =0x=0, and it has a power of2, so the multiplicity of this zero is2.

STEP 7

The factors x−11x - \sqrt{11}x−11​ and x+11x + \sqrt{11}x+11​ correspond to the zeros x=11x = \sqrt{11}x=11​ and x=−11x = -\sqrt{11}x=−11​, respectively, and they each have a power of1, so the multiplicities of these zeros are1.

STEP 8

The graph of a polynomial function crosses the x-axis at a zero if the zero has an odd multiplicity, and touches the x-axis and turns around at a zero if the zero has an even multiplicity.

STEP 9

So, the graph of fff crosses the x-axis at the zeros x=11x = \sqrt{11}x=11​ and x=−11x = -\sqrt{11}x=−11​, and touches the x-axis and turns around at the zero x=x =x=.

STEP 10

The maximum number of turning points of a polynomial function is one less than the degree of the function. The degree of f(x)=−5x2(x2−)f(x)=-5x^{2}(x^{2}-)f(x)=−5x2(x2−) is4, so the maximum number of turning points is4 - =3.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionDetermine the end behavior of $f(x)=-5 x^{2}(x^{2}-11)$ and find its real zeros and their multiplicities.

Assumptions1. The function is given by $f(x)=-5x^{}(x^{}-11)$ . We are asked to find the real zeros of the function, their multiplicities, the behavior of the graph at the x-intercepts, and the maximum number of turning points.

To find the real zeros of the function, we need to set the function equal to zero and solve for $x$ .
$-5x^{2}(x^{2}-11) =0$

We can solve this equation by setting each factor equal to zero.
$-5x^{2} =0 \quad \text{and} \quad x^{2}-11 =0$

olving the first equation gives us one of the real zeros.
$-x^{2} =0 \implies x =0$

olving the second equation gives us the other two real zeros.
$x^{2}-11 =0 \implies x = \sqrt{11}, -\sqrt{11}$

The multiplicities of the zeros are the powers of the corresponding factors in the factored form of the polynomial. The factor $x^{2}$ corresponds to the zero $x =0$ , and it has a power of2, so the multiplicity of this zero is2.

The factors $x - \sqrt{11}$ and $x + \sqrt{11}$ correspond to the zeros $x = \sqrt{11}$ and $x = -\sqrt{11}$ , respectively, and they each have a power of1, so the multiplicities of these zeros are1.

So, the graph of $f$ crosses the x-axis at the zeros $x = \sqrt{11}$ and $x = -\sqrt{11}$ , and touches the x-axis and turns around at the zero $x =$ .

The maximum number of turning points of a polynomial function is one less than the degree of the function. The degree of $f(x)=-5x^{2}(x^{2}-)$ is4, so the maximum number of turning points is4 - =3.