QuestionSketch the graph of the piecewise function $f(x)=\begin{cases} 4+x, & x \leq 2 \\ x^{2}+2, & x>2 \end{cases}$ .

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is defined as a piecewise function with two parts. . The first part of the function is $4+x$ for $x \leq$ .
3. The second part of the function is $x^{}+$ for $x >$ .

STEP 2

We will start by sketching the first part of the function, $4+x$ for $x \leq2$ . This is a linear function with a slope of1 and a y-intercept of4.

STEP 3

To sketch this part of the function, we can start by plotting the y-intercept at $(0,)$ .

STEP 4

Then, since the slope is1, we can go up one unit and to the right one unit from the y-intercept to find another point on the line. Continue this process until you reach $x =2$ .

STEP 5

At $x =2$ , the value of the function is $4+2=$ . So, we have the point $(2,)$ on the graph.

STEP 6

Since the function is defined for $x \leq2$ in this part, we will draw a filled circle at $(2,6)$ to indicate that this point is included in the graph.

STEP 7

Now, we will sketch the second part of the function, $x^{2}+2$ for $x >2$ . This is a quadratic function with a vertex at $(0,2)$ .

STEP 8

However, since this part of the function is only defined for $x >2$ , we will only sketch the part of the parabola that is to the right of $x =2$ .

STEP 9

At $x =2$ , the value of the function is $2^{2}+2=6$ . So, we have the point $(2,6)$ on the graph.

STEP 10

Since the function is defined for $x >2$ in this part, we will draw an open circle at $(2,6)$ to indicate that this point is not included in the graph.

STEP 11

Now, we can sketch the rest of the parabola for $x >$ .

STEP 12

The graph of the function $f(x)$ is now complete. It consists of a line segment for $x \leq2$ and a part of a parabola for $x >2$ .

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QuestionSketch the graph of the piecewise function $f(x)=\begin{cases} 4+x, & x \leq 2 \\ x^{2}+2, & x>2 \end{cases}$ .

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is defined as a piecewise function with two parts. . The first part of the function is $4+x$ for $x \leq$ .
3. The second part of the function is $x^{}+$ for $x >$ .

STEP 2

We will start by sketching the first part of the function, $4+x$ for $x \leq2$ . This is a linear function with a slope of1 and a y-intercept of4.

STEP 3

To sketch this part of the function, we can start by plotting the y-intercept at $(0,)$ .

STEP 4

Then, since the slope is1, we can go up one unit and to the right one unit from the y-intercept to find another point on the line. Continue this process until you reach $x =2$ .

STEP 5

At $x =2$ , the value of the function is $4+2=$ . So, we have the point $(2,)$ on the graph.

STEP 6

Since the function is defined for $x \leq2$ in this part, we will draw a filled circle at $(2,6)$ to indicate that this point is included in the graph.

STEP 7

Now, we will sketch the second part of the function, $x^{2}+2$ for $x >2$ . This is a quadratic function with a vertex at $(0,2)$ .

STEP 8

However, since this part of the function is only defined for $x >2$ , we will only sketch the part of the parabola that is to the right of $x =2$ .

STEP 9

At $x =2$ , the value of the function is $2^{2}+2=6$ . So, we have the point $(2,6)$ on the graph.

STEP 10

Since the function is defined for $x >2$ in this part, we will draw an open circle at $(2,6)$ to indicate that this point is not included in the graph.

STEP 11

Now, we can sketch the rest of the parabola for $x >$ .

STEP 12

The graph of the function $f(x)$ is now complete. It consists of a line segment for $x \leq2$ and a part of a parabola for $x >2$ .

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QuestionSketch the graph of the piecewise function f(x)={4+x,x≤2x2+2,x>2f(x)=\begin{cases} 4+x, & x \leq 2 \\ x^{2}+2, & x>2 \end{cases}f(x)={4+x,x2+2,​x≤2x>2​.

Studdy Solution

STEP 1

Assumptions1. The function f(x)f(x)f(x) is defined as a piecewise function with two parts. . The first part of the function is 4+x4+x4+x for x≤x \leqx≤.3. The second part of the function is x+x^{}+x+ for x>x >x>.

STEP 2

We will start by sketching the first part of the function, 4+x4+x4+x for x≤2x \leq2x≤2. This is a linear function with a slope of1 and a y-intercept of4.

STEP 3

To sketch this part of the function, we can start by plotting the y-intercept at (0,)(0,)(0,).

STEP 4

Then, since the slope is1, we can go up one unit and to the right one unit from the y-intercept to find another point on the line. Continue this process until you reach x=2x =2x=2.

STEP 5

At x=2x =2x=2, the value of the function is 4+2=4+2=4+2=. So, we have the point (2,)(2,)(2,) on the graph.

STEP 6

Since the function is defined for x≤2x \leq2x≤2 in this part, we will draw a filled circle at (2,6)(2,6)(2,6) to indicate that this point is included in the graph.

STEP 7

Now, we will sketch the second part of the function, x2+2x^{2}+2x2+2 for x>2x >2x>2. This is a quadratic function with a vertex at (0,2)(0,2)(0,2).

STEP 8

However, since this part of the function is only defined for x>2x >2x>2, we will only sketch the part of the parabola that is to the right of x=2x =2x=2.

STEP 9

At x=2x =2x=2, the value of the function is 22+2=62^{2}+2=622+2=6. So, we have the point (2,6)(2,6)(2,6) on the graph.

STEP 10

Since the function is defined for x>2x >2x>2 in this part, we will draw an open circle at (2,6)(2,6)(2,6) to indicate that this point is not included in the graph.

STEP 11

Now, we can sketch the rest of the parabola for x>x >x>.

STEP 12

The graph of the function f(x)f(x)f(x) is now complete. It consists of a line segment for x≤2x \leq2x≤2 and a part of a parabola for x>2x >2x>2.

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QuestionSketch the graph of the piecewise function $f(x)=\begin{cases} 4+x, & x \leq 2 \\ x^{2}+2, & x>2 \end{cases}$ .

Assumptions1. The function $f(x)$ is defined as a piecewise function with two parts. . The first part of the function is $4+x$ for $x \leq$ .
3. The second part of the function is $x^{}+$ for $x >$ .

We will start by sketching the first part of the function, $4+x$ for $x \leq2$ . This is a linear function with a slope of1 and a y-intercept of4.

To sketch this part of the function, we can start by plotting the y-intercept at $(0,)$ .

Then, since the slope is1, we can go up one unit and to the right one unit from the y-intercept to find another point on the line. Continue this process until you reach $x =2$ .

At $x =2$ , the value of the function is $4+2=$ . So, we have the point $(2,)$ on the graph.

Since the function is defined for $x \leq2$ in this part, we will draw a filled circle at $(2,6)$ to indicate that this point is included in the graph.

Now, we will sketch the second part of the function, $x^{2}+2$ for $x >2$ . This is a quadratic function with a vertex at $(0,2)$ .

However, since this part of the function is only defined for $x >2$ , we will only sketch the part of the parabola that is to the right of $x =2$ .

At $x =2$ , the value of the function is $2^{2}+2=6$ . So, we have the point $(2,6)$ on the graph.

Since the function is defined for $x >2$ in this part, we will draw an open circle at $(2,6)$ to indicate that this point is not included in the graph.

Now, we can sketch the rest of the parabola for $x >$ .

The graph of the function $f(x)$ is now complete. It consists of a line segment for $x \leq2$ and a part of a parabola for $x >2$ .