QuestionAnalyze the polynomial $f(x)=(x+8)(x-1)^{2}$ :
(a) End behavior: behaves like $y=x^{3}$ for large $|x|$ . (b) Intercepts: $x$ -intercepts are $-8, 1$ ; $y$ -intercept is 8. (c) Zeros: $-8, 1$ ; check if graph crosses/touches $x$ -axis. (d) Maximum turning points: (whole number).

Studdy Solution

STEP 1

Assumptions1. The polynomial function is $f(x)=(x+8)(x-1)^{}$ . We need to analyze the function in terms of end behavior, intercepts, zeros and turning points

STEP 2

To determine the end behavior of the graph of the function, we need to look at the highest degree term in the polynomial. This term will dominate the behavior of the function for large values of $|x|$ .

STEP 3

Expand the polynomial to identify the highest degree term.
$f(x) = (x+8)(x-1)^{2} = (x+8)(x^{2}-2x+1)$

STEP 4

Further expand the polynomial.
$f(x) = x^{3} -2x^{2} + x +8x^{2} -16x +8 = x^{3} +6x^{2} -15x +8$

STEP 5

The highest degree term in the polynomial is $x^{3}$ . Therefore, the end behavior of the graph of the function is like $y=x^{3}$ for large values of $|x|$ .

STEP 6

To find the $x$ -intercepts of the graph of the function, we need to set $f(x) =0$ and solve for $x$ .

STEP 7

Set the function equal to zero and solve for $x$ .
$0 = (x+)(x-1)^{2}$

STEP 8

The solutions to this equation are the $x$ -intercepts of the graph of the function. The $x$ -intercepts are $x = -8$ and $x =1$ .

STEP 9

To find the $y$ -intercept of the graph of the function, we need to set $x =$ and solve for $y$ .

STEP 10

Set $x =0$ in the function and solve for $y$ .
$y = f(0) = (0+8)(0-)^{2} =8$

STEP 11

The $y$ -intercept of the graph of the function is $y =8$ .

STEP 12

The zeros of the function are the values of $x$ for which $f(x) =0$ . These are also the $x$ -intercepts of the graph of the function. The zeros of the function are $x = -8$ and $x =$ .

STEP 13

The multiplicity of a zero is the number of times it appears as a root of the polynomial. In this case, $x = -8$ appears once and $x =$ appears twice. Therefore, the zero $x = -8$ has multiplicity and the zero $x =$ has multiplicity2.

STEP 14

If the multiplicity of a zero is odd, the graph crosses the $x$ -axis at that point. If the multiplicity is even, the graph touches the $x$ -axis at that point and turns around. Therefore, the graph crosses the $x$ -axis at $x = -8$ and touches the $x$ -axis at $x =$ .

STEP 15

The maximum number of turning points on the graph of a polynomial function is one less than the degree of the polynomial. The degree of the polynomial $f(x) = x^{3} +x^{2} -15x +8$ is3.

STEP 16

Therefore, the maximum number of turning points on the graph of the function is $3 - =2$ .

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QuestionAnalyze the polynomial $f(x)=(x+8)(x-1)^{2}$ :
(a) End behavior: behaves like $y=x^{3}$ for large $|x|$ . (b) Intercepts: $x$ -intercepts are $-8, 1$ ; $y$ -intercept is 8. (c) Zeros: $-8, 1$ ; check if graph crosses/touches $x$ -axis. (d) Maximum turning points: (whole number).

Studdy Solution

STEP 1

Assumptions1. The polynomial function is $f(x)=(x+8)(x-1)^{}$ . We need to analyze the function in terms of end behavior, intercepts, zeros and turning points

STEP 2

To determine the end behavior of the graph of the function, we need to look at the highest degree term in the polynomial. This term will dominate the behavior of the function for large values of $|x|$ .

STEP 3

Expand the polynomial to identify the highest degree term.
$f(x) = (x+8)(x-1)^{2} = (x+8)(x^{2}-2x+1)$

STEP 4

Further expand the polynomial.
$f(x) = x^{3} -2x^{2} + x +8x^{2} -16x +8 = x^{3} +6x^{2} -15x +8$

STEP 5

The highest degree term in the polynomial is $x^{3}$ . Therefore, the end behavior of the graph of the function is like $y=x^{3}$ for large values of $|x|$ .

STEP 6

To find the $x$ -intercepts of the graph of the function, we need to set $f(x) =0$ and solve for $x$ .

STEP 7

Set the function equal to zero and solve for $x$ .
$0 = (x+)(x-1)^{2}$

STEP 8

The solutions to this equation are the $x$ -intercepts of the graph of the function. The $x$ -intercepts are $x = -8$ and $x =1$ .

STEP 9

To find the $y$ -intercept of the graph of the function, we need to set $x =$ and solve for $y$ .

STEP 10

Set $x =0$ in the function and solve for $y$ .
$y = f(0) = (0+8)(0-)^{2} =8$

STEP 11

The $y$ -intercept of the graph of the function is $y =8$ .

STEP 12

The zeros of the function are the values of $x$ for which $f(x) =0$ . These are also the $x$ -intercepts of the graph of the function. The zeros of the function are $x = -8$ and $x =$ .

STEP 13

The multiplicity of a zero is the number of times it appears as a root of the polynomial. In this case, $x = -8$ appears once and $x =$ appears twice. Therefore, the zero $x = -8$ has multiplicity and the zero $x =$ has multiplicity2.

STEP 14

If the multiplicity of a zero is odd, the graph crosses the $x$ -axis at that point. If the multiplicity is even, the graph touches the $x$ -axis at that point and turns around. Therefore, the graph crosses the $x$ -axis at $x = -8$ and touches the $x$ -axis at $x =$ .

STEP 15

The maximum number of turning points on the graph of a polynomial function is one less than the degree of the polynomial. The degree of the polynomial $f(x) = x^{3} +x^{2} -15x +8$ is3.

STEP 16

Therefore, the maximum number of turning points on the graph of the function is $3 - =2$ .

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Studdy Solution

STEP 1

Assumptions1. The polynomial function is f(x)=(x+8)(x−1)f(x)=(x+8)(x-1)^{}f(x)=(x+8)(x−1) . We need to analyze the function in terms of end behavior, intercepts, zeros and turning points

STEP 2

To determine the end behavior of the graph of the function, we need to look at the highest degree term in the polynomial. This term will dominate the behavior of the function for large values of ∣x∣|x|∣x∣.

STEP 3

Expand the polynomial to identify the highest degree term.f(x)=(x+8)(x−1)2=(x+8)(x2−2x+1)f(x) = (x+8)(x-1)^{2} = (x+8)(x^{2}-2x+1)f(x)=(x+8)(x−1)2=(x+8)(x2−2x+1)

STEP 4

Further expand the polynomial.f(x)=x3−2x2+x+8x2−16x+8=x3+6x2−15x+8f(x) = x^{3} -2x^{2} + x +8x^{2} -16x +8 = x^{3} +6x^{2} -15x +8f(x)=x3−2x2+x+8x2−16x+8=x3+6x2−15x+8

STEP 5

The highest degree term in the polynomial is x3x^{3}x3. Therefore, the end behavior of the graph of the function is like y=x3y=x^{3}y=x3 for large values of ∣x∣|x|∣x∣.

STEP 6

To find the xxx-intercepts of the graph of the function, we need to set f(x)=0f(x) =0f(x)=0 and solve for xxx.

STEP 7

Set the function equal to zero and solve for xxx.0=(x+)(x−1)20 = (x+)(x-1)^{2}0=(x+)(x−1)2

STEP 8

The solutions to this equation are the xxx-intercepts of the graph of the function. The xxx-intercepts are x=−8x = -8x=−8 and x=1x =1x=1.

STEP 9

To find the yyy-intercept of the graph of the function, we need to set x=x =x= and solve for yyy.

STEP 10

Set x=0x =0x=0 in the function and solve for yyy.y=f(0)=(0+8)(0−)2=8y = f(0) = (0+8)(0-)^{2} =8y=f(0)=(0+8)(0−)2=8

STEP 11

The yyy-intercept of the graph of the function is y=8y =8y=8.

STEP 12

The zeros of the function are the values of xxx for which f(x)=0f(x) =0f(x)=0. These are also the xxx-intercepts of the graph of the function. The zeros of the function are x=−8x = -8x=−8 and x=x =x=.

STEP 13

The multiplicity of a zero is the number of times it appears as a root of the polynomial. In this case, x=−8x = -8x=−8 appears once and x=x =x= appears twice. Therefore, the zero x=−8x = -8x=−8 has multiplicity and the zero x=x =x= has multiplicity2.

STEP 14

If the multiplicity of a zero is odd, the graph crosses the xxx-axis at that point. If the multiplicity is even, the graph touches the xxx-axis at that point and turns around. Therefore, the graph crosses the xxx-axis at x=−8x = -8x=−8 and touches the xxx-axis at x=x =x=.

STEP 15

The maximum number of turning points on the graph of a polynomial function is one less than the degree of the polynomial. The degree of the polynomial f(x)=x3+x2−15x+8f(x) = x^{3} +x^{2} -15x +8f(x)=x3+x2−15x+8 is3.

STEP 16

Therefore, the maximum number of turning points on the graph of the function is 3−=23 - =23−=2.

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Assumptions1. The polynomial function is $f(x)=(x+8)(x-1)^{}$ . We need to analyze the function in terms of end behavior, intercepts, zeros and turning points

To determine the end behavior of the graph of the function, we need to look at the highest degree term in the polynomial. This term will dominate the behavior of the function for large values of $|x|$ .

Expand the polynomial to identify the highest degree term.
$f(x) = (x+8)(x-1)^{2} = (x+8)(x^{2}-2x+1)$

Further expand the polynomial.
$f(x) = x^{3} -2x^{2} + x +8x^{2} -16x +8 = x^{3} +6x^{2} -15x +8$

The highest degree term in the polynomial is $x^{3}$ . Therefore, the end behavior of the graph of the function is like $y=x^{3}$ for large values of $|x|$ .

To find the $x$ -intercepts of the graph of the function, we need to set $f(x) =0$ and solve for $x$ .

Set the function equal to zero and solve for $x$ .
$0 = (x+)(x-1)^{2}$

The solutions to this equation are the $x$ -intercepts of the graph of the function. The $x$ -intercepts are $x = -8$ and $x =1$ .

To find the $y$ -intercept of the graph of the function, we need to set $x =$ and solve for $y$ .

Set $x =0$ in the function and solve for $y$ .
$y = f(0) = (0+8)(0-)^{2} =8$

The $y$ -intercept of the graph of the function is $y =8$ .

The zeros of the function are the values of $x$ for which $f(x) =0$ . These are also the $x$ -intercepts of the graph of the function. The zeros of the function are $x = -8$ and $x =$ .

The multiplicity of a zero is the number of times it appears as a root of the polynomial. In this case, $x = -8$ appears once and $x =$ appears twice. Therefore, the zero $x = -8$ has multiplicity and the zero $x =$ has multiplicity2.

If the multiplicity of a zero is odd, the graph crosses the $x$ -axis at that point. If the multiplicity is even, the graph touches the $x$ -axis at that point and turns around. Therefore, the graph crosses the $x$ -axis at $x = -8$ and touches the $x$ -axis at $x =$ .

The maximum number of turning points on the graph of a polynomial function is one less than the degree of the polynomial. The degree of the polynomial $f(x) = x^{3} +x^{2} -15x +8$ is3.

Therefore, the maximum number of turning points on the graph of the function is $3 - =2$ .