QuestionMatch each function $h$ with its transformation: (a) $h(x)=f(x)+c$ , (b) $h(x)=f(x)-c$ , (c) $h(x)=f(x+c)$ , (d) $h(x)=f(x-c)$ .

Studdy Solution

STEP 1

Assumptions1. The functions $h(x)$ are transformations of a base function $f(x)$ . . The constant $c$ is positive.
3. The transformations can be either horizontal or vertical shifts.

STEP 2

Let's start with the first function $h(x)=f(x)+c$ .
This function represents a shift of the base function $f(x)$ , but we need to determine whether it's a horizontal or vertical shift, and in which direction.

STEP 3

The transformation $h(x)=f(x)+c$ adds a constant $c$ to the output of the function $f(x)$ .
This means that for any input $x$ , the output of the function $h(x)$ is $c$ units higher than the output of the function $f(x)$ .

STEP 4

Therefore, the function $h(x)=f(x)+c$ represents a vertical shift of $f$ , $c$ units up.

STEP 5

Next, let's analyze the function $h(x)=f(x)-c$ .
This function subtracts a constant $c$ from the output of the function $f(x)$ .
This means that for any input $x$ , the output of the function $h(x)$ is $c$ units lower than the output of the function $f(x)$ .

STEP 6

Therefore, the function $h(x)=f(x)-c$ represents a vertical shift of $f$ , $c$ units down.

STEP 7

Now, let's analyze the function $h(x)=f(x+c)$ .
This function changes the input to the function $f(x)$ by adding a constant $c$ .
This means that for any input $x$ , the function $h(x)$ evaluates $f$ at a point $c$ units to the left of $x$ .

STEP 8

Therefore, the function $h(x)=f(x+c)$ represents a horizontal shift of $f$ , $c$ units to the left.

STEP 9

Finally, let's analyze the function $h(x)=f(x-c)$ .
This function changes the input to the function $f(x)$ by subtracting a constant $c$ .
This means that for any input $x$ , the function $h(x)$ evaluates $f$ at a point $c$ units to the right of $x$ .

STEP 10

Therefore, the function $h(x)=f(x-c)$ represents a horizontal shift of $f$ , $c$ units to the right.
So, the transformations are as follows(a) $h(x)=f(x)+c$ represents a vertical shift of $f, c$ units up. (b) $h(x)=f(x)-c$ represents a vertical shift of $f, c$ units down. (c) $h(x)=f(x+c)$ represents a horizontal shift of $f, c$ units to the left. (d) $h(x)=f(x-c)$ represents a horizontal shift of $f, c$ units to the right.

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QuestionMatch each function $h$ with its transformation: (a) $h(x)=f(x)+c$ , (b) $h(x)=f(x)-c$ , (c) $h(x)=f(x+c)$ , (d) $h(x)=f(x-c)$ .

Studdy Solution

STEP 1

Assumptions1. The functions $h(x)$ are transformations of a base function $f(x)$ . . The constant $c$ is positive.
3. The transformations can be either horizontal or vertical shifts.

STEP 2

Let's start with the first function $h(x)=f(x)+c$ .
This function represents a shift of the base function $f(x)$ , but we need to determine whether it's a horizontal or vertical shift, and in which direction.

STEP 3

The transformation $h(x)=f(x)+c$ adds a constant $c$ to the output of the function $f(x)$ .
This means that for any input $x$ , the output of the function $h(x)$ is $c$ units higher than the output of the function $f(x)$ .

STEP 4

Therefore, the function $h(x)=f(x)+c$ represents a vertical shift of $f$ , $c$ units up.

STEP 5

Next, let's analyze the function $h(x)=f(x)-c$ .
This function subtracts a constant $c$ from the output of the function $f(x)$ .
This means that for any input $x$ , the output of the function $h(x)$ is $c$ units lower than the output of the function $f(x)$ .

STEP 6

Therefore, the function $h(x)=f(x)-c$ represents a vertical shift of $f$ , $c$ units down.

STEP 7

Now, let's analyze the function $h(x)=f(x+c)$ .
This function changes the input to the function $f(x)$ by adding a constant $c$ .
This means that for any input $x$ , the function $h(x)$ evaluates $f$ at a point $c$ units to the left of $x$ .

STEP 8

Therefore, the function $h(x)=f(x+c)$ represents a horizontal shift of $f$ , $c$ units to the left.

STEP 9

Finally, let's analyze the function $h(x)=f(x-c)$ .
This function changes the input to the function $f(x)$ by subtracting a constant $c$ .
This means that for any input $x$ , the function $h(x)$ evaluates $f$ at a point $c$ units to the right of $x$ .

STEP 10

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QuestionMatch each function hhh with its transformation: (a) h(x)=f(x)+ch(x)=f(x)+ch(x)=f(x)+c, (b) h(x)=f(x)−ch(x)=f(x)-ch(x)=f(x)−c, (c) h(x)=f(x+c)h(x)=f(x+c)h(x)=f(x+c), (d) h(x)=f(x−c)h(x)=f(x-c)h(x)=f(x−c).

Studdy Solution

STEP 1

Assumptions1. The functions h(x)h(x)h(x) are transformations of a base function f(x)f(x)f(x). . The constant ccc is positive.3. The transformations can be either horizontal or vertical shifts.

STEP 2

Let's start with the first function h(x)=f(x)+ch(x)=f(x)+ch(x)=f(x)+c.This function represents a shift of the base function f(x)f(x)f(x), but we need to determine whether it's a horizontal or vertical shift, and in which direction.

STEP 3

The transformation h(x)=f(x)+ch(x)=f(x)+ch(x)=f(x)+c adds a constant ccc to the output of the function f(x)f(x)f(x).This means that for any input xxx, the output of the function h(x)h(x)h(x) is ccc units higher than the output of the function f(x)f(x)f(x).

STEP 4

Therefore, the function h(x)=f(x)+ch(x)=f(x)+ch(x)=f(x)+c represents a vertical shift of fff, ccc units up.

STEP 5

Next, let's analyze the function h(x)=f(x)−ch(x)=f(x)-ch(x)=f(x)−c.This function subtracts a constant ccc from the output of the function f(x)f(x)f(x).This means that for any input xxx, the output of the function h(x)h(x)h(x) is ccc units lower than the output of the function f(x)f(x)f(x).

STEP 6

Therefore, the function h(x)=f(x)−ch(x)=f(x)-ch(x)=f(x)−c represents a vertical shift of fff, ccc units down.

STEP 7

Now, let's analyze the function h(x)=f(x+c)h(x)=f(x+c)h(x)=f(x+c).This function changes the input to the function f(x)f(x)f(x) by adding a constant ccc.This means that for any input xxx, the function h(x)h(x)h(x) evaluates fff at a point ccc units to the left of xxx.

STEP 8

Therefore, the function h(x)=f(x+c)h(x)=f(x+c)h(x)=f(x+c) represents a horizontal shift of fff, ccc units to the left.

STEP 9

Finally, let's analyze the function h(x)=f(x−c)h(x)=f(x-c)h(x)=f(x−c).This function changes the input to the function f(x)f(x)f(x) by subtracting a constant ccc.This means that for any input xxx, the function h(x)h(x)h(x) evaluates fff at a point ccc units to the right of xxx.

STEP 10

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QuestionMatch each function $h$ with its transformation: (a) $h(x)=f(x)+c$ , (b) $h(x)=f(x)-c$ , (c) $h(x)=f(x+c)$ , (d) $h(x)=f(x-c)$ .

Assumptions1. The functions $h(x)$ are transformations of a base function $f(x)$ . . The constant $c$ is positive.
3. The transformations can be either horizontal or vertical shifts.

Let's start with the first function $h(x)=f(x)+c$ .
This function represents a shift of the base function $f(x)$ , but we need to determine whether it's a horizontal or vertical shift, and in which direction.

The transformation $h(x)=f(x)+c$ adds a constant $c$ to the output of the function $f(x)$ .
This means that for any input $x$ , the output of the function $h(x)$ is $c$ units higher than the output of the function $f(x)$ .

Therefore, the function $h(x)=f(x)+c$ represents a vertical shift of $f$ , $c$ units up.

Next, let's analyze the function $h(x)=f(x)-c$ .
This function subtracts a constant $c$ from the output of the function $f(x)$ .
This means that for any input $x$ , the output of the function $h(x)$ is $c$ units lower than the output of the function $f(x)$ .

Therefore, the function $h(x)=f(x)-c$ represents a vertical shift of $f$ , $c$ units down.

Now, let's analyze the function $h(x)=f(x+c)$ .
This function changes the input to the function $f(x)$ by adding a constant $c$ .
This means that for any input $x$ , the function $h(x)$ evaluates $f$ at a point $c$ units to the left of $x$ .

Therefore, the function $h(x)=f(x+c)$ represents a horizontal shift of $f$ , $c$ units to the left.

Finally, let's analyze the function $h(x)=f(x-c)$ .
This function changes the input to the function $f(x)$ by subtracting a constant $c$ .
This means that for any input $x$ , the function $h(x)$ evaluates $f$ at a point $c$ units to the right of $x$ .