QuestionFind the starting point of the function $f(x)=\sqrt{x+2}-2$ on the coordinate plane. $([?],[])$

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=\sqrt{x+}-$ . We are looking for the ordered pair where the function begins on the coordinate plane. This is the point where the function is first defined.

STEP 2

The function $f(x)$ is defined only when the value under the square root is non-negative, as we cannot take the square root of a negative number in the real number system.So, we need to find the smallest value of $x$ for which $x+2 \geq0$ .

STEP 3

olving the inequality $x+2 \geq0$ gives us the smallest value of $x$ for which the function is defined.
$x+2 \geq0$

STEP 4

Subtract2 from both sides of the inequality to isolate $x$ .
$x \geq -2$

STEP 5

The smallest value of $x$ for which the function is defined is $x=-2$ .

STEP 6

Now, we substitute $x=-2$ into the function $f(x)$ to find the corresponding $y$ -coordinate.
$f(-2) = \sqrt{-2+2}-2$

STEP 7

implify the expression under the square root.
$f(-2) = \sqrt{0}-2$

STEP 8

Calculate the square root of0.
$f(-2) =0-2$

STEP 9

Subtract2 from to find the $y$ -coordinate.
$f(-2) = -2$

STEP 10

The ordered pair where the function $f(x)=\sqrt{x+2}-2$ begins on the coordinate plane is $(-2, -2)$ .
$(-2,-2)$

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QuestionFind the starting point of the function $f(x)=\sqrt{x+2}-2$ on the coordinate plane. $([?],[])$

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=\sqrt{x+}-$ . We are looking for the ordered pair where the function begins on the coordinate plane. This is the point where the function is first defined.

STEP 2

The function $f(x)$ is defined only when the value under the square root is non-negative, as we cannot take the square root of a negative number in the real number system.So, we need to find the smallest value of $x$ for which $x+2 \geq0$ .

STEP 3

olving the inequality $x+2 \geq0$ gives us the smallest value of $x$ for which the function is defined.
$x+2 \geq0$

STEP 4

Subtract2 from both sides of the inequality to isolate $x$ .
$x \geq -2$

STEP 5

The smallest value of $x$ for which the function is defined is $x=-2$ .

STEP 6

Now, we substitute $x=-2$ into the function $f(x)$ to find the corresponding $y$ -coordinate.
$f(-2) = \sqrt{-2+2}-2$

STEP 7

implify the expression under the square root.
$f(-2) = \sqrt{0}-2$

STEP 8

Calculate the square root of0.
$f(-2) =0-2$

STEP 9

Subtract2 from to find the $y$ -coordinate.
$f(-2) = -2$

STEP 10

The ordered pair where the function $f(x)=\sqrt{x+2}-2$ begins on the coordinate plane is $(-2, -2)$ .
$(-2,-2)$

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QuestionFind the starting point of the function f(x)=x+2−2f(x)=\sqrt{x+2}-2f(x)=x+2​−2 on the coordinate plane. ([?],[])([?],[])([?],[])

Studdy Solution

STEP 1

Assumptions1. The function is f(x)=x+−f(x)=\sqrt{x+}-f(x)=x+​− . We are looking for the ordered pair where the function begins on the coordinate plane. This is the point where the function is first defined.

STEP 2

The function f(x)f(x)f(x) is defined only when the value under the square root is non-negative, as we cannot take the square root of a negative number in the real number system.So, we need to find the smallest value of xxx for which x+2≥0x+2 \geq0x+2≥0.

STEP 3

olving the inequality x+2≥0x+2 \geq0x+2≥0 gives us the smallest value of xxx for which the function is defined.x+2≥0x+2 \geq0x+2≥0

STEP 4

Subtract2 from both sides of the inequality to isolate xxx.x≥−2x \geq -2x≥−2

STEP 5

The smallest value of xxx for which the function is defined is x=−2x=-2x=−2.

STEP 6

Now, we substitute x=−2x=-2x=−2 into the function f(x)f(x)f(x) to find the corresponding yyy-coordinate.f(−2)=−2+2−2f(-2) = \sqrt{-2+2}-2f(−2)=−2+2​−2

STEP 7

implify the expression under the square root.f(−2)=0−2f(-2) = \sqrt{0}-2f(−2)=0​−2

STEP 8

Calculate the square root of0.f(−2)=0−2f(-2) =0-2f(−2)=0−2

STEP 9

Subtract2 from to find the yyy-coordinate.f(−2)=−2f(-2) = -2f(−2)=−2

STEP 10

The ordered pair where the function f(x)=x+2−2f(x)=\sqrt{x+2}-2f(x)=x+2​−2 begins on the coordinate plane is (−2,−2)(-2, -2)(−2,−2).(−2,−2)(-2,-2) (−2,−2)

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QuestionFind the starting point of the function $f(x)=\sqrt{x+2}-2$ on the coordinate plane. $([?],[])$

Assumptions1. The function is $f(x)=\sqrt{x+}-$ . We are looking for the ordered pair where the function begins on the coordinate plane. This is the point where the function is first defined.

The function $f(x)$ is defined only when the value under the square root is non-negative, as we cannot take the square root of a negative number in the real number system.So, we need to find the smallest value of $x$ for which $x+2 \geq0$ .

olving the inequality $x+2 \geq0$ gives us the smallest value of $x$ for which the function is defined.
$x+2 \geq0$

Subtract2 from both sides of the inequality to isolate $x$ .
$x \geq -2$

The smallest value of $x$ for which the function is defined is $x=-2$ .

Now, we substitute $x=-2$ into the function $f(x)$ to find the corresponding $y$ -coordinate.
$f(-2) = \sqrt{-2+2}-2$

implify the expression under the square root.
$f(-2) = \sqrt{0}-2$

Calculate the square root of0.
$f(-2) =0-2$

Subtract2 from to find the $y$ -coordinate.
$f(-2) = -2$

The ordered pair where the function $f(x)=\sqrt{x+2}-2$ begins on the coordinate plane is $(-2, -2)$ .
$(-2,-2)$