Math / Algebra

QuestionDetermine the range of the function $y=\sqrt{x-2}$ . $\{y \in R ; y \geq 0\}$ $\{y \in R\}$ $\{x \in R ; x \geq-2\}$ $\{x \in R\}$
Question 2 (1 point)

Studdy Solution

STEP 1

What is this asking? What are all the possible output values ( $y$ ) we can get from this square root function? Watch out! Remember that we can't take the square root of a negative number in real-number math!

STEP 2

1. Figure out the allowed $x$ values
2. See how $x$ values affect $y$ values
3. Determine the range

STEP 3

Alright, let's start by looking at what's inside that square root: $x - 2$ .
We know that whatever is under a square root has to be greater than or equal to zero.
So, $x - 2 \geq 0$ .

STEP 4

To isolate $x$ , we'll add 2 to both sides of the inequality.
This gives us $x - 2 + 2 \geq 0 + 2$ , which simplifies to $x \geq 2$ .
This means $x$ can be any number 2 or greater!

STEP 5

Now, let's see what happens to $y$ when we plug in different values of $x$ .
If we plug in our smallest possible $x$ value, which is $x = 2$ , we get $y = \sqrt{2 - 2} = \sqrt{0} = 0$ .
So, our smallest $y$ value is zero!

STEP 6

What happens as $x$ gets bigger?
Well, as $x$ increases, $x - 2$ also increases.
And since the square root function is an increasing function, $y = \sqrt{x - 2}$ will also increase.

STEP 7

Let's try a bigger $x$ value, like $x = 6$ .
We get $y = \sqrt{6 - 2} = \sqrt{4} = 2$ .
See? $y$ got bigger too!
In fact, $y$ can get as big as we want it to, as long as we keep making $x$ bigger.

STEP 8

We've figured out that the smallest $y$ value is 0, and $y$ can get as large as we want.
This means the range of our function is all real numbers greater than or equal to zero.
We can write this as $\{y \in \mathbb{R} ; y \geq 0\}$ .

STEP 9

The range of the function $y = \sqrt{x - 2}$ is $\{y \in \mathbb{R} ; y \geq 0\}$ .

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QuestionDetermine the range of the function $y=\sqrt{x-2}$ . $\{y \in R ; y \geq 0\}$ $\{y \in R\}$ $\{x \in R ; x \geq-2\}$ $\{x \in R\}$
Question 2 (1 point)

Studdy Solution

STEP 1

What is this asking? What are all the possible output values ( $y$ ) we can get from this square root function? Watch out! Remember that we can't take the square root of a negative number in real-number math!

STEP 2

1. Figure out the allowed $x$ values
2. See how $x$ values affect $y$ values
3. Determine the range

STEP 3

Alright, let's start by looking at what's inside that square root: $x - 2$ .
We know that whatever is under a square root has to be greater than or equal to zero.
So, $x - 2 \geq 0$ .

STEP 4

To isolate $x$ , we'll add 2 to both sides of the inequality.
This gives us $x - 2 + 2 \geq 0 + 2$ , which simplifies to $x \geq 2$ .
This means $x$ can be any number 2 or greater!

STEP 5

Now, let's see what happens to $y$ when we plug in different values of $x$ .
If we plug in our smallest possible $x$ value, which is $x = 2$ , we get $y = \sqrt{2 - 2} = \sqrt{0} = 0$ .
So, our smallest $y$ value is zero!

STEP 6

What happens as $x$ gets bigger?
Well, as $x$ increases, $x - 2$ also increases.
And since the square root function is an increasing function, $y = \sqrt{x - 2}$ will also increase.

STEP 7

Let's try a bigger $x$ value, like $x = 6$ .
We get $y = \sqrt{6 - 2} = \sqrt{4} = 2$ .
See? $y$ got bigger too!
In fact, $y$ can get as big as we want it to, as long as we keep making $x$ bigger.

STEP 8

We've figured out that the smallest $y$ value is 0, and $y$ can get as large as we want.
This means the range of our function is all real numbers greater than or equal to zero.
We can write this as $\{y \in \mathbb{R} ; y \geq 0\}$ .

STEP 9

The range of the function $y = \sqrt{x - 2}$ is $\{y \in \mathbb{R} ; y \geq 0\}$ .

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QuestionDetermine the range of the function y=x−2y=\sqrt{x-2}y=x−2​. {y∈R;y≥0}\{y \in R ; y \geq 0\}{y∈R;y≥0} {y∈R}\{y \in R\}{y∈R} {x∈R;x≥−2}\{x \in R ; x \geq-2\}{x∈R;x≥−2} {x∈R}\{x \in R\}{x∈R}Question 2 (1 point)

Studdy Solution

STEP 1

What is this asking? What are all the possible output values (yyy) we can get from this square root function? Watch out! Remember that we can't take the square root of a negative number in real-number math!

STEP 2

1. Figure out the allowed xxx values2. See how xxx values affect yyy values3. Determine the range

STEP 3

Alright, let's **start** by looking at what's *inside* that square root: x−2x - 2x−2.We know that whatever is under a square root has to be greater than or equal to zero.So, x−2≥0x - 2 \geq 0x−2≥0.

STEP 4

To **isolate** xxx, we'll add 2 to both sides of the inequality.This gives us x−2+2≥0+2x - 2 + 2 \geq 0 + 2x−2+2≥0+2, which simplifies to x≥2x \geq 2x≥2.This means xxx can be **any number** 2 or greater!

STEP 5

Now, let's see what happens to yyy when we plug in different values of xxx.If we plug in our **smallest possible** xxx value, which is x=2x = 2x=2, we get y=2−2=0=0y = \sqrt{2 - 2} = \sqrt{0} = 0y=2−2​=0​=0.So, our **smallest** yyy value is **zero**!

STEP 6

What happens as xxx gets bigger?Well, as xxx increases, x−2x - 2x−2 also increases.And since the square root function is an *increasing* function, y=x−2y = \sqrt{x - 2}y=x−2​ will also increase.

STEP 7

Let's try a **bigger** xxx value, like x=6x = 6x=6.We get y=6−2=4=2y = \sqrt{6 - 2} = \sqrt{4} = 2y=6−2​=4​=2.See? yyy got bigger too!In fact, yyy can get as **big** as we want it to, as long as we keep making xxx bigger.

STEP 8

We've figured out that the **smallest** yyy value is 0, and yyy can get as **large** as we want.This means the **range** of our function is all real numbers greater than or equal to zero.We can write this as {y∈R;y≥0}\{y \in \mathbb{R} ; y \geq 0\}{y∈R;y≥0}.

STEP 9

The range of the function y=x−2y = \sqrt{x - 2}y=x−2​ is {y∈R;y≥0}\{y \in \mathbb{R} ; y \geq 0\}{y∈R;y≥0}.

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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.

QuestionDetermine the range of the function $y=\sqrt{x-2}$ . $\{y \in R ; y \geq 0\}$ $\{y \in R\}$ $\{x \in R ; x \geq-2\}$ $\{x \in R\}$
Question 2 (1 point)

What is this asking? What are all the possible output values ( $y$ ) we can get from this square root function? Watch out! Remember that we can't take the square root of a negative number in real-number math!

1. Figure out the allowed $x$ values
2. See how $x$ values affect $y$ values
3. Determine the range

Alright, let's start by looking at what's inside that square root: $x - 2$ .
We know that whatever is under a square root has to be greater than or equal to zero.
So, $x - 2 \geq 0$ .

To isolate $x$ , we'll add 2 to both sides of the inequality.
This gives us $x - 2 + 2 \geq 0 + 2$ , which simplifies to $x \geq 2$ .
This means $x$ can be any number 2 or greater!

Now, let's see what happens to $y$ when we plug in different values of $x$ .
If we plug in our smallest possible $x$ value, which is $x = 2$ , we get $y = \sqrt{2 - 2} = \sqrt{0} = 0$ .
So, our smallest $y$ value is zero!

What happens as $x$ gets bigger?
Well, as $x$ increases, $x - 2$ also increases.
And since the square root function is an increasing function, $y = \sqrt{x - 2}$ will also increase.

Let's try a bigger $x$ value, like $x = 6$ .
We get $y = \sqrt{6 - 2} = \sqrt{4} = 2$ .
See? $y$ got bigger too!
In fact, $y$ can get as big as we want it to, as long as we keep making $x$ bigger.

We've figured out that the smallest $y$ value is 0, and $y$ can get as large as we want.
This means the range of our function is all real numbers greater than or equal to zero.
We can write this as $\{y \in \mathbb{R} ; y \geq 0\}$ .

The range of the function $y = \sqrt{x - 2}$ is $\{y \in \mathbb{R} ; y \geq 0\}$ .