Solved on Oct 31, 2023

Find the $x$ -intercepts and zeros of $f(x)=(x+2)^{2}(x-4)^{3}(x-3)$ . Express the intercepts and zeros as ordered pairs.

STEP 1

Assumptions1. The function is given by $f(x)=(x+)^{}(x-4)^{3}(x-3)$ . We need to find the x-intercepts at which the function crosses the axis3. We need to find the zeros at which the function "flattens out"

STEP 2

The x-intercepts of a function are the points where the function crosses the x-axis. This happens when the function value $f(x)$ is equal to zero. So, we need to solve the equation $f(x) =0$ to find the x-intercepts.
$f(x)=(x+2)^{2}(x-4)^{}(x-) =0$

STEP 3

The equation $f(x) =0$ is a product of three factors. A product is zero if and only if at least one of the factors is zero. So, we can set each factor equal to zero and solve for $x$ .
$(x+2)^{2} =0, (x-)^{3} =0, (x-3) =0$

STEP 4

olving the equations $(x+2)^{2} =0$ , $(x-4)^{3} =0$ , and $(x-3) =0$ for $x$ gives the x-intercepts.
$x = -2, x =4, x =3$

STEP 5

The x-intercepts are the points where the function crosses the x-axis. These points are given by the ordered pairs $(x,0)$ , where $x$ is the x-coordinate of the intercept. So, the x-intercepts are $(-2,0)$ , $(4,0)$ , and $(3,0)$ .

STEP 6

The zeros at which the function "flattens out" are the points where the derivative of the function is zero. The derivative of a function gives the slope of the function at a given point. When the derivative is zero, the function has a horizontal tangent line, which means it "flattens out". So, we need to find the derivative of the function and set it equal to zero.
$f'(x) =0$

STEP 7

To find the derivative of the function $f(x)=(x+2)^{2}(x-4)^{3}(x-3)$ , we need to use the product rule and the chain rule. The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

STEP 8

First, let's find the derivative of each factor in the function.
(x+2)^{2}' =2(x+2)(x-4)^{3}' =3(x-4)^{2} $(x-3)' =1$

STEP 9

Now, let's apply the product rule to find the derivative of the function $f(x)=(x+2)^{2}(x-4)^{3}(x-3)$ .
f'(x) = (x+2)^{2}'(x-4)^{3}(x-3) + (x+2)^{2}(x-4)^{3}'(x-3) + (x+2)^{2}(x-4)^{3}(x-3)'

STEP 10

Substitute the derivatives of each factor into the equation for $f'(x)$ .
$f'(x) =2(x+2)(x-4)^{3}(x-3) + (x+2)^{2}3(x-4)^{2}(x-3) + (x+2)^{2}(x-4)^{3}$

STEP 11

Set the derivative equal to zero and solve for $x$ .
$(x+)(x-4)^{3}(x-3) + (x+)^{}3(x-4)^{}(x-3) + (x+)^{}(x-4)^{3} =0$

STEP 12

This equation is quite complex to solve analytically. However, we can notice that the x-intercepts $x = -2,4,$ are also solutions of this equation. This is because at these points, one of the factors in each term of the equation becomes zero, making the whole equation equal to zero.

STEP 13

The zeros at which the function "flattens out" are the points where the derivative of the function is zero. These points are given by the ordered pairs $(x, f(x))$ , where $x$ is the x-coordinate of the zero. So, the zeros at which the function "flattens out" are $(-2, f(-2))$ , $(, f())$ , and $(3, f(3))$ .
The x-intercepts at which $f$ crosses the axis are $(-2,0)$ , $(,0)$ , and $(3,0)$ . The zeros at which $f$ "flattens out" are $(-2, f(-2))$ , $(, f())$ , and $(3, f(3))$ .

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Find the $x$ -intercepts and zeros of $f(x)=(x+2)^{2}(x-4)^{3}(x-3)$ . Express the intercepts and zeros as ordered pairs.

STEP 1

Assumptions1. The function is given by $f(x)=(x+)^{}(x-4)^{3}(x-3)$ . We need to find the x-intercepts at which the function crosses the axis3. We need to find the zeros at which the function "flattens out"

STEP 2

The x-intercepts of a function are the points where the function crosses the x-axis. This happens when the function value $f(x)$ is equal to zero. So, we need to solve the equation $f(x) =0$ to find the x-intercepts.
$f(x)=(x+2)^{2}(x-4)^{}(x-) =0$

STEP 3

The equation $f(x) =0$ is a product of three factors. A product is zero if and only if at least one of the factors is zero. So, we can set each factor equal to zero and solve for $x$ .
$(x+2)^{2} =0, (x-)^{3} =0, (x-3) =0$

STEP 4

olving the equations $(x+2)^{2} =0$ , $(x-4)^{3} =0$ , and $(x-3) =0$ for $x$ gives the x-intercepts.
$x = -2, x =4, x =3$

STEP 5

The x-intercepts are the points where the function crosses the x-axis. These points are given by the ordered pairs $(x,0)$ , where $x$ is the x-coordinate of the intercept. So, the x-intercepts are $(-2,0)$ , $(4,0)$ , and $(3,0)$ .

STEP 6

STEP 7

STEP 8

First, let's find the derivative of each factor in the function.
(x+2)^{2}' =2(x+2)(x-4)^{3}' =3(x-4)^{2} $(x-3)' =1$

STEP 9

Now, let's apply the product rule to find the derivative of the function $f(x)=(x+2)^{2}(x-4)^{3}(x-3)$ .
f'(x) = (x+2)^{2}'(x-4)^{3}(x-3) + (x+2)^{2}(x-4)^{3}'(x-3) + (x+2)^{2}(x-4)^{3}(x-3)'

STEP 10

Substitute the derivatives of each factor into the equation for $f'(x)$ .
$f'(x) =2(x+2)(x-4)^{3}(x-3) + (x+2)^{2}3(x-4)^{2}(x-3) + (x+2)^{2}(x-4)^{3}$

STEP 11

Set the derivative equal to zero and solve for $x$ .
$(x+)(x-4)^{3}(x-3) + (x+)^{}3(x-4)^{}(x-3) + (x+)^{}(x-4)^{3} =0$

STEP 12

This equation is quite complex to solve analytically. However, we can notice that the x-intercepts $x = -2,4,$ are also solutions of this equation. This is because at these points, one of the factors in each term of the equation becomes zero, making the whole equation equal to zero.

STEP 13

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Find the xxx-intercepts and zeros of f(x)=(x+2)2(x−4)3(x−3)f(x)=(x+2)^{2}(x-4)^{3}(x-3)f(x)=(x+2)2(x−4)3(x−3). Express the intercepts and zeros as ordered pairs.

STEP 1

Assumptions1. The function is given by f(x)=(x+)(x−4)3(x−3)f(x)=(x+)^{}(x-4)^{3}(x-3)f(x)=(x+)(x−4)3(x−3). We need to find the x-intercepts at which the function crosses the axis3. We need to find the zeros at which the function "flattens out"

STEP 2

STEP 3

The equation f(x)=0f(x) =0f(x)=0 is a product of three factors. A product is zero if and only if at least one of the factors is zero. So, we can set each factor equal to zero and solve for xxx.(x+2)2=0,(x−)3=0,(x−3)=0(x+2)^{2} =0, (x-)^{3} =0, (x-3) =0(x+2)2=0,(x−)3=0,(x−3)=0

STEP 4

olving the equations (x+2)2=0(x+2)^{2} =0(x+2)2=0, (x−4)3=0(x-4)^{3} =0(x−4)3=0, and (x−3)=0(x-3) =0(x−3)=0 for xxx gives the x-intercepts.x=−2,x=4,x=3x = -2, x =4, x =3x=−2,x=4,x=3

STEP 5

The x-intercepts are the points where the function crosses the x-axis. These points are given by the ordered pairs (x,0)(x,0)(x,0), where xxx is the x-coordinate of the intercept. So, the x-intercepts are (−2,0)(-2,0)(−2,0), (4,0)(4,0)(4,0), and (3,0)(3,0)(3,0).

STEP 6

STEP 7

STEP 8

First, let's find the derivative of each factor in the function.(x+2)^{2}' =2(x+2)(x-4)^{3}' =3(x-4)^{2}(x−3)′=1(x-3)' =1(x−3)′=1

STEP 9

Now, let's apply the product rule to find the derivative of the function f(x)=(x+2)2(x−4)3(x−3)f(x)=(x+2)^{2}(x-4)^{3}(x-3)f(x)=(x+2)2(x−4)3(x−3).f'(x) = (x+2)^{2}'(x-4)^{3}(x-3) + (x+2)^{2}(x-4)^{3}'(x-3) + (x+2)^{2}(x-4)^{3}(x-3)'

STEP 10

Substitute the derivatives of each factor into the equation for f′(x)f'(x)f′(x).f′(x)=2(x+2)(x−4)3(x−3)+(x+2)23(x−4)2(x−3)+(x+2)2(x−4)3f'(x) =2(x+2)(x-4)^{3}(x-3) + (x+2)^{2}3(x-4)^{2}(x-3) + (x+2)^{2}(x-4)^{3}f′(x)=2(x+2)(x−4)3(x−3)+(x+2)23(x−4)2(x−3)+(x+2)2(x−4)3

STEP 11

Set the derivative equal to zero and solve for xxx.(x+)(x−4)3(x−3)+(x+)3(x−4)(x−3)+(x+)(x−4)3=0(x+)(x-4)^{3}(x-3) + (x+)^{}3(x-4)^{}(x-3) + (x+)^{}(x-4)^{3} =0(x+)(x−4)3(x−3)+(x+)3(x−4)(x−3)+(x+)(x−4)3=0

STEP 12

This equation is quite complex to solve analytically. However, we can notice that the x-intercepts x=−2,4,x = -2,4,x=−2,4, are also solutions of this equation. This is because at these points, one of the factors in each term of the equation becomes zero, making the whole equation equal to zero.

STEP 13

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Find the $x$ -intercepts and zeros of $f(x)=(x+2)^{2}(x-4)^{3}(x-3)$ . Express the intercepts and zeros as ordered pairs.

Assumptions1. The function is given by $f(x)=(x+)^{}(x-4)^{3}(x-3)$ . We need to find the x-intercepts at which the function crosses the axis3. We need to find the zeros at which the function "flattens out"

The equation $f(x) =0$ is a product of three factors. A product is zero if and only if at least one of the factors is zero. So, we can set each factor equal to zero and solve for $x$ .
$(x+2)^{2} =0, (x-)^{3} =0, (x-3) =0$

olving the equations $(x+2)^{2} =0$ , $(x-4)^{3} =0$ , and $(x-3) =0$ for $x$ gives the x-intercepts.
$x = -2, x =4, x =3$

The x-intercepts are the points where the function crosses the x-axis. These points are given by the ordered pairs $(x,0)$ , where $x$ is the x-coordinate of the intercept. So, the x-intercepts are $(-2,0)$ , $(4,0)$ , and $(3,0)$ .

First, let's find the derivative of each factor in the function.
(x+2)^{2}' =2(x+2)(x-4)^{3}' =3(x-4)^{2} $(x-3)' =1$

Now, let's apply the product rule to find the derivative of the function $f(x)=(x+2)^{2}(x-4)^{3}(x-3)$ .
f'(x) = (x+2)^{2}'(x-4)^{3}(x-3) + (x+2)^{2}(x-4)^{3}'(x-3) + (x+2)^{2}(x-4)^{3}(x-3)'

Substitute the derivatives of each factor into the equation for $f'(x)$ .
$f'(x) =2(x+2)(x-4)^{3}(x-3) + (x+2)^{2}3(x-4)^{2}(x-3) + (x+2)^{2}(x-4)^{3}$

Set the derivative equal to zero and solve for $x$ .
$(x+)(x-4)^{3}(x-3) + (x+)^{}3(x-4)^{}(x-3) + (x+)^{}(x-4)^{3} =0$

This equation is quite complex to solve analytically. However, we can notice that the x-intercepts $x = -2,4,$ are also solutions of this equation. This is because at these points, one of the factors in each term of the equation becomes zero, making the whole equation equal to zero.