Question Choose the graph of the piecewise function $f(x) = \begin{cases} 3x-2, & x \leq 1 \\ x+2, & x > 1 \end{cases}$ .

Studdy Solution

STEP 1

Assumptions
1. The function is a piecewise function, which means it is defined by different expressions for different intervals of the domain.
2. The function is defined by the expression $3x - 2$ for $x \leq 1$ .
3. The function is defined by the expression $x + 2$ for $x > 1$ .

STEP 2

We need to graph the function for each piece separately. Let's start with the first piece $3x - 2$ for $x \leq 1$ .

STEP 3

To graph $3x - 2$ for $x \leq 1$ , we need to plot the line $y = 3x - 2$ for all $x$ values less than or equal to 1.

STEP 4

Let's find the y-coordinate when $x = 1$ for the first piece. Substitute $x = 1$ into the equation $y = 3x - 2$ .
$y = 3(1) - 2$

STEP 5

Calculate the y-coordinate.
$y = 3(1) - 2 = 1$

STEP 6

So, the point $(1,1)$ is on the graph of the first piece of the function.

STEP 7

Now, let's graph the second piece $x + 2$ for $x > 1$ .

STEP 8

To graph $x + 2$ for $x > 1$ , we need to plot the line $y = x + 2$ for all $x$ values greater than 1.

STEP 9

Let's find the y-coordinate when $x = 1$ for the second piece. Substitute $x = 1$ into the equation $y = x + 2$ .
$y = 1 + 2$

STEP 10

Calculate the y-coordinate.
$y = 1 + 2 = 3$

STEP 11

So, the point $(1,3)$ is on the graph of the second piece of the function.

STEP 12

However, since $x > 1$ for the second piece, the point $(1,3)$ is not included in the graph. We indicate this by drawing an open circle at $(1,3)$ .

STEP 13

Now, we can draw the graph of the piecewise function. For $x \leq 1$ , we draw the line $y = 3x - 2$ up to the point $(1,1)$ . For $x > 1$ , we draw the line $y = x + 2$ starting from the open circle at $(1,3)$ .

Was this helpful?

Question Choose the graph of the piecewise function $f(x) = \begin{cases} 3x-2, & x \leq 1 \\ x+2, & x > 1 \end{cases}$ .

Studdy Solution

STEP 1

Assumptions
1. The function is a piecewise function, which means it is defined by different expressions for different intervals of the domain.
2. The function is defined by the expression $3x - 2$ for $x \leq 1$ .
3. The function is defined by the expression $x + 2$ for $x > 1$ .

STEP 2

We need to graph the function for each piece separately. Let's start with the first piece $3x - 2$ for $x \leq 1$ .

STEP 3

To graph $3x - 2$ for $x \leq 1$ , we need to plot the line $y = 3x - 2$ for all $x$ values less than or equal to 1.

STEP 4

Let's find the y-coordinate when $x = 1$ for the first piece. Substitute $x = 1$ into the equation $y = 3x - 2$ .
$y = 3(1) - 2$

STEP 5

Calculate the y-coordinate.
$y = 3(1) - 2 = 1$

STEP 6

So, the point $(1,1)$ is on the graph of the first piece of the function.

STEP 7

Now, let's graph the second piece $x + 2$ for $x > 1$ .

STEP 8

To graph $x + 2$ for $x > 1$ , we need to plot the line $y = x + 2$ for all $x$ values greater than 1.

STEP 9

Let's find the y-coordinate when $x = 1$ for the second piece. Substitute $x = 1$ into the equation $y = x + 2$ .
$y = 1 + 2$

STEP 10

Calculate the y-coordinate.
$y = 1 + 2 = 3$

STEP 11

So, the point $(1,3)$ is on the graph of the second piece of the function.

STEP 12

However, since $x > 1$ for the second piece, the point $(1,3)$ is not included in the graph. We indicate this by drawing an open circle at $(1,3)$ .

STEP 13

Now, we can draw the graph of the piecewise function. For $x \leq 1$ , we draw the line $y = 3x - 2$ up to the point $(1,1)$ . For $x > 1$ , we draw the line $y = x + 2$ starting from the open circle at $(1,3)$ .

Get Studdy

Studdy solves anything!

Start learning now

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

Question Choose the graph of the piecewise function f(x)={3x−2,x≤1x+2,x>1f(x) = \begin{cases} 3x-2, & x \leq 1 \\ x+2, & x > 1 \end{cases}f(x)={3x−2,x+2,​x≤1x>1​.

Studdy Solution

STEP 1

STEP 2

We need to graph the function for each piece separately. Let's start with the first piece 3x−23x - 23x−2 for x≤1x \leq 1x≤1.

STEP 3

To graph 3x−23x - 23x−2 for x≤1x \leq 1x≤1, we need to plot the line y=3x−2y = 3x - 2y=3x−2 for all xxx values less than or equal to 1.

STEP 4

Let's find the y-coordinate when x=1x = 1x=1 for the first piece. Substitute x=1x = 1x=1 into the equation y=3x−2y = 3x - 2y=3x−2.y=3(1)−2y = 3(1) - 2y=3(1)−2

STEP 5

Calculate the y-coordinate.y=3(1)−2=1y = 3(1) - 2 = 1y=3(1)−2=1

STEP 6

So, the point (1,1)(1,1)(1,1) is on the graph of the first piece of the function.

STEP 7

Now, let's graph the second piece x+2x + 2x+2 for x>1x > 1x>1.

STEP 8

To graph x+2x + 2x+2 for x>1x > 1x>1, we need to plot the line y=x+2y = x + 2y=x+2 for all xxx values greater than 1.

STEP 9

Let's find the y-coordinate when x=1x = 1x=1 for the second piece. Substitute x=1x = 1x=1 into the equation y=x+2y = x + 2y=x+2.y=1+2y = 1 + 2y=1+2

STEP 10

Calculate the y-coordinate.y=1+2=3y = 1 + 2 = 3y=1+2=3

STEP 11

So, the point (1,3)(1,3)(1,3) is on the graph of the second piece of the function.

STEP 12

However, since x>1x > 1x>1 for the second piece, the point (1,3)(1,3)(1,3) is not included in the graph. We indicate this by drawing an open circle at (1,3)(1,3)(1,3).

STEP 13

Now, we can draw the graph of the piecewise function. For x≤1x \leq 1x≤1, we draw the line y=3x−2y = 3x - 2y=3x−2 up to the point (1,1)(1,1)(1,1). For x>1x > 1x>1, we draw the line y=x+2y = x + 2y=x+2 starting from the open circle at (1,3)(1,3)(1,3).

Get Studdy

Studdy solves anything!

Start learning now

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

Question Choose the graph of the piecewise function $f(x) = \begin{cases} 3x-2, & x \leq 1 \\ x+2, & x > 1 \end{cases}$ .

We need to graph the function for each piece separately. Let's start with the first piece $3x - 2$ for $x \leq 1$ .

To graph $3x - 2$ for $x \leq 1$ , we need to plot the line $y = 3x - 2$ for all $x$ values less than or equal to 1.

Let's find the y-coordinate when $x = 1$ for the first piece. Substitute $x = 1$ into the equation $y = 3x - 2$ .
$y = 3(1) - 2$

Calculate the y-coordinate.
$y = 3(1) - 2 = 1$

So, the point $(1,1)$ is on the graph of the first piece of the function.

Now, let's graph the second piece $x + 2$ for $x > 1$ .

To graph $x + 2$ for $x > 1$ , we need to plot the line $y = x + 2$ for all $x$ values greater than 1.

Let's find the y-coordinate when $x = 1$ for the second piece. Substitute $x = 1$ into the equation $y = x + 2$ .
$y = 1 + 2$

Calculate the y-coordinate.
$y = 1 + 2 = 3$

So, the point $(1,3)$ is on the graph of the second piece of the function.

However, since $x > 1$ for the second piece, the point $(1,3)$ is not included in the graph. We indicate this by drawing an open circle at $(1,3)$ .

Now, we can draw the graph of the piecewise function. For $x \leq 1$ , we draw the line $y = 3x - 2$ up to the point $(1,1)$ . For $x > 1$ , we draw the line $y = x + 2$ starting from the open circle at $(1,3)$ .