QuestionWhich point is on the graph of the piecewise function $f(x)$ defined as: $f(x)=-9$ for $x<-6$ , $f(x)=1$ for $x=-6$ , and $f(x)=7x-1$ for $x>-6$ ? Options: $(-6,-9)$ , $(-6,-43)$ , $(1,-6)$ , $(0,-9)$ , $(-5,-36)$ , or None.

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is defined as a piecewise function with three parts - $f(x) = -9$ for $x < -6$ - $f(x) =1$ for $x = -6$ - $f(x) =7x -1$ for $x > -6$
. We are given the points $(-6,-9)$ , $(-6,-43)$ , $(1,-6)$ , $(0,-9)$ , $(-5,-36)$ , and "None of the above" as possible points on the graph of $f(x)$ .

STEP 2

We will check each point by plugging the x-values into the appropriate part of the piecewise function and seeing if we get the corresponding y-value.

STEP 3

First, let's check the point $(-6,-9)$ .
For $x=-6$ , the function $f(x)$ is defined as $f(x)=1$ . So, we should have $f(-6) =1$ , not $-9$ . So, $(-6,-9)$ is not on the graph of $f(x)$ .

STEP 4

Next, let's check the point $(-6,-43)$ .
Again, for $x=-6$ , the function $f(x)$ is defined as $f(x)=1$ . So, we should have $f(-6) =1$ , not $-43$ . So, $(-6,-43)$ is not on the graph of $f(x)$ .

STEP 5

Now, let's check the point $(1,-)$ .
For $x=1$ , the function $f(x)$ is defined as $f(x)=7x-1$ . So, we should have $f(1) =7(1) -1 =$ , not $-$ . So, $(1,-)$ is not on the graph of $f(x)$ .

STEP 6

Next, let's check the point $(0,-9)$ .
For $x=0$ , the function $f(x)$ is defined as $f(x)=x-1$ . So, we should have $f(0) =(0) -1 = -1$ , not $-9$ . So, $(0,-9)$ is not on the graph of $f(x)$ .

STEP 7

Finally, let's check the point $(-5,-36)$ .
For $x=-5$ , the function $f(x)$ is defined as $f(x)=7x-1$ . So, we should have $f(-5) =7(-5) -1 = -36$ . So, $(-5,-36)$ is on the graph of $f(x)$ .
The point on the graph of $f(x)$ is $(-5,-36)$ .

Was this helpful?

QuestionWhich point is on the graph of the piecewise function $f(x)$ defined as: $f(x)=-9$ for $x<-6$ , $f(x)=1$ for $x=-6$ , and $f(x)=7x-1$ for $x>-6$ ? Options: $(-6,-9)$ , $(-6,-43)$ , $(1,-6)$ , $(0,-9)$ , $(-5,-36)$ , or None.

Studdy Solution

STEP 1

STEP 2

We will check each point by plugging the x-values into the appropriate part of the piecewise function and seeing if we get the corresponding y-value.

STEP 3

First, let's check the point $(-6,-9)$ .
For $x=-6$ , the function $f(x)$ is defined as $f(x)=1$ . So, we should have $f(-6) =1$ , not $-9$ . So, $(-6,-9)$ is not on the graph of $f(x)$ .

STEP 4

Next, let's check the point $(-6,-43)$ .
Again, for $x=-6$ , the function $f(x)$ is defined as $f(x)=1$ . So, we should have $f(-6) =1$ , not $-43$ . So, $(-6,-43)$ is not on the graph of $f(x)$ .

STEP 5

Now, let's check the point $(1,-)$ .
For $x=1$ , the function $f(x)$ is defined as $f(x)=7x-1$ . So, we should have $f(1) =7(1) -1 =$ , not $-$ . So, $(1,-)$ is not on the graph of $f(x)$ .

STEP 6

Next, let's check the point $(0,-9)$ .
For $x=0$ , the function $f(x)$ is defined as $f(x)=x-1$ . So, we should have $f(0) =(0) -1 = -1$ , not $-9$ . So, $(0,-9)$ is not on the graph of $f(x)$ .

STEP 7

Finally, let's check the point $(-5,-36)$ .
For $x=-5$ , the function $f(x)$ is defined as $f(x)=7x-1$ . So, we should have $f(-5) =7(-5) -1 = -36$ . So, $(-5,-36)$ is on the graph of $f(x)$ .
The point on the graph of $f(x)$ is $(-5,-36)$ .

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.

Studdy Solution

STEP 1

STEP 2

We will check each point by plugging the x-values into the appropriate part of the piecewise function and seeing if we get the corresponding y-value.

STEP 3

First, let's check the point (−6,−9)(-6,-9)(−6,−9).For x=−6x=-6x=−6, the function f(x)f(x)f(x) is defined as f(x)=1f(x)=1f(x)=1. So, we should have f(−6)=1f(-6) =1f(−6)=1, not −9-9−9. So, (−6,−9)(-6,-9)(−6,−9) is not on the graph of f(x)f(x)f(x).

STEP 4

Next, let's check the point (−6,−43)(-6,-43)(−6,−43).Again, for x=−6x=-6x=−6, the function f(x)f(x)f(x) is defined as f(x)=1f(x)=1f(x)=1. So, we should have f(−6)=1f(-6) =1f(−6)=1, not −43-43−43. So, (−6,−43)(-6,-43)(−6,−43) is not on the graph of f(x)f(x)f(x).

STEP 5

Now, let's check the point (1,−)(1,-)(1,−).For x=1x=1x=1, the function f(x)f(x)f(x) is defined as f(x)=7x−1f(x)=7x-1f(x)=7x−1. So, we should have f(1)=7(1)−1=f(1) =7(1) -1 =f(1)=7(1)−1=, not −-−. So, (1,−)(1,-)(1,−) is not on the graph of f(x)f(x)f(x).

STEP 6

Next, let's check the point (0,−9)(0,-9)(0,−9).For x=0x=0x=0, the function f(x)f(x)f(x) is defined as f(x)=x−1f(x)=x-1f(x)=x−1. So, we should have f(0)=(0)−1=−1f(0) =(0) -1 = -1f(0)=(0)−1=−1, not −9-9−9. So, (0,−9)(0,-9)(0,−9) is not on the graph of f(x)f(x)f(x).

STEP 7

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.

First, let's check the point $(-6,-9)$ .
For $x=-6$ , the function $f(x)$ is defined as $f(x)=1$ . So, we should have $f(-6) =1$ , not $-9$ . So, $(-6,-9)$ is not on the graph of $f(x)$ .

Next, let's check the point $(-6,-43)$ .
Again, for $x=-6$ , the function $f(x)$ is defined as $f(x)=1$ . So, we should have $f(-6) =1$ , not $-43$ . So, $(-6,-43)$ is not on the graph of $f(x)$ .

Now, let's check the point $(1,-)$ .
For $x=1$ , the function $f(x)$ is defined as $f(x)=7x-1$ . So, we should have $f(1) =7(1) -1 =$ , not $-$ . So, $(1,-)$ is not on the graph of $f(x)$ .

Next, let's check the point $(0,-9)$ .
For $x=0$ , the function $f(x)$ is defined as $f(x)=x-1$ . So, we should have $f(0) =(0) -1 = -1$ , not $-9$ . So, $(0,-9)$ is not on the graph of $f(x)$ .