Math / Algebra

QuestionGraph the function. $f(x)=3 \sqrt{x-2}$
Plot four points on the graph of the function: the le

Studdy Solution

STEP 1

What is this asking? We need to draw a smooth curve of $f(x)=3\sqrt{x-2}$ by finding some smart points that will help us sketch it properly. Watch out! Remember the domain starts at $x=2$ (not zero) because we can't take the square root of a negative number!

STEP 2

1. Find the domain
2. Find key points
3. Determine shape
4. Plot and connect

STEP 3

Let's think about what values of $x$ we can actually use here.
We've got a square root, and we know we can't take the square root of negative numbers.
So what's inside our square root needs to be positive or zero.
$x-2 \geq 0$

STEP 4

Solving this is super simple - just add 2 to both sides:
$x \geq 2$
This means our graph can only exist for values of $x$ that are 2 or greater.
That's our domain!

STEP 5

Let's start with the easiest point - when $x=2$ .
This is our starting point because it's where the domain begins:
$f(2) = 3\sqrt{2-2} = 3\sqrt{0} = 3 \cdot 0 = \mathbf{0}$
So our first point is $(2,0)$ !

STEP 6

Let's find some more nice points.
When $x=3$ :
$f(3) = 3\sqrt{3-2} = 3\sqrt{1} = 3 \cdot 1 = \mathbf{3}$
We've got $(3,3)$ !

STEP 7

When $x=6$ :
$f(6) = 3\sqrt{6-2} = 3\sqrt{4} = 3 \cdot 2 = \mathbf{6}$
Adding $(6,6)$ to our collection.

STEP 8

One more nice point, let's try $x=11$ :
$f(11) = 3\sqrt{11-2} = 3\sqrt{9} = 3 \cdot 3 = \mathbf{9}$
And we've got $(11,9)$ !

STEP 9

This is a square root function, but it's been modified in two ways: - It's been shifted 2 units right (because of the -2 inside) - It's been stretched vertically by a factor of 3 (because of the 3 outside)

STEP 10

The function will start at our point $(2,0)$ and curve up and to the right, just like a regular square root function, but steeper because of that factor of 3.

STEP 11

Now we can plot our points: - $(2,0)$ - Where we start - $(3,3)$ - Our first nice point - $(6,6)$ - Another helpful point - $(11,9)$ - One more to show the curve flattening out

STEP 12

Connect these points with a smooth curve, making sure it curves exactly like a square root function should - steeper at the start and gradually becoming less steep.

STEP 13

The graph is a smooth curve starting at point $(2,0)$ , passing through $(3,3)$ , $(6,6)$ , and $(11,9)$ , extending infinitely upward and to the right, but only existing for $x \geq 2$ .

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QuestionGraph the function. $f(x)=3 \sqrt{x-2}$
Plot four points on the graph of the function: the le

Studdy Solution

STEP 1

What is this asking? We need to draw a smooth curve of $f(x)=3\sqrt{x-2}$ by finding some smart points that will help us sketch it properly. Watch out! Remember the domain starts at $x=2$ (not zero) because we can't take the square root of a negative number!

STEP 2

1. Find the domain
2. Find key points
3. Determine shape
4. Plot and connect

STEP 3

Let's think about what values of $x$ we can actually use here.
We've got a square root, and we know we can't take the square root of negative numbers.
So what's inside our square root needs to be positive or zero.
$x-2 \geq 0$

STEP 4

Solving this is super simple - just add 2 to both sides:
$x \geq 2$
This means our graph can only exist for values of $x$ that are 2 or greater.
That's our domain!

STEP 5

Let's start with the easiest point - when $x=2$ .
This is our starting point because it's where the domain begins:
$f(2) = 3\sqrt{2-2} = 3\sqrt{0} = 3 \cdot 0 = \mathbf{0}$
So our first point is $(2,0)$ !

STEP 6

Let's find some more nice points.
When $x=3$ :
$f(3) = 3\sqrt{3-2} = 3\sqrt{1} = 3 \cdot 1 = \mathbf{3}$
We've got $(3,3)$ !

STEP 7

When $x=6$ :
$f(6) = 3\sqrt{6-2} = 3\sqrt{4} = 3 \cdot 2 = \mathbf{6}$
Adding $(6,6)$ to our collection.

STEP 8

One more nice point, let's try $x=11$ :
$f(11) = 3\sqrt{11-2} = 3\sqrt{9} = 3 \cdot 3 = \mathbf{9}$
And we've got $(11,9)$ !

STEP 9

This is a square root function, but it's been modified in two ways: - It's been shifted 2 units right (because of the -2 inside) - It's been stretched vertically by a factor of 3 (because of the 3 outside)

STEP 10

The function will start at our point $(2,0)$ and curve up and to the right, just like a regular square root function, but steeper because of that factor of 3.

STEP 11

Now we can plot our points: - $(2,0)$ - Where we start - $(3,3)$ - Our first nice point - $(6,6)$ - Another helpful point - $(11,9)$ - One more to show the curve flattening out

STEP 12

Connect these points with a smooth curve, making sure it curves exactly like a square root function should - steeper at the start and gradually becoming less steep.

STEP 13

The graph is a smooth curve starting at point $(2,0)$ , passing through $(3,3)$ , $(6,6)$ , and $(11,9)$ , extending infinitely upward and to the right, but only existing for $x \geq 2$ .

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QuestionGraph the function. f(x)=3x−2f(x)=3 \sqrt{x-2}f(x)=3x−2​Plot four points on the graph of the function: the le

Studdy Solution

STEP 1

What is this asking? We need to draw a smooth curve of f(x)=3x−2f(x)=3\sqrt{x-2}f(x)=3x−2​ by finding some smart points that will help us sketch it properly. Watch out! Remember the domain starts at x=2x=2x=2 (not zero) because we can't take the square root of a negative number!

STEP 2

1. Find the domain2. Find key points3. Determine shape4. Plot and connect

STEP 3

Let's think about what values of xxx we can actually use here.We've got a square root, and we know we can't take the square root of negative numbers.So what's inside our square root needs to be positive or zero.x−2≥0x-2 \geq 0x−2≥0

STEP 4

Solving this is super simple - just add 2 to both sides:x≥2x \geq 2x≥2This means our graph can only exist for values of xxx that are **2 or greater**.That's our domain!

STEP 5

Let's start with the easiest point - when x=2x=2x=2.This is our starting point because it's where the domain begins:f(2)=32−2=30=3⋅0=0f(2) = 3\sqrt{2-2} = 3\sqrt{0} = 3 \cdot 0 = \mathbf{0}f(2)=32−2​=30​=3⋅0=0So our first point is (2,0)(2,0)(2,0)!

STEP 6

Let's find some more nice points.When x=3x=3x=3:f(3)=33−2=31=3⋅1=3f(3) = 3\sqrt{3-2} = 3\sqrt{1} = 3 \cdot 1 = \mathbf{3}f(3)=33−2​=31​=3⋅1=3We've got (3,3)(3,3)(3,3)!

STEP 7

When x=6x=6x=6:f(6)=36−2=34=3⋅2=6f(6) = 3\sqrt{6-2} = 3\sqrt{4} = 3 \cdot 2 = \mathbf{6}f(6)=36−2​=34​=3⋅2=6Adding (6,6)(6,6)(6,6) to our collection.

STEP 8

One more nice point, let's try x=11x=11x=11:f(11)=311−2=39=3⋅3=9f(11) = 3\sqrt{11-2} = 3\sqrt{9} = 3 \cdot 3 = \mathbf{9}f(11)=311−2​=39​=3⋅3=9And we've got (11,9)(11,9)(11,9)!

STEP 9

This is a square root function, but it's been modified in two ways: - It's been shifted **2 units right** (because of the -2 inside) - It's been stretched **vertically by a factor of 3** (because of the 3 outside)

STEP 10

The function will start at our point (2,0)(2,0)(2,0) and curve up and to the right, just like a regular square root function, but steeper because of that factor of 3.

STEP 11

Now we can plot our points: - (2,0)(2,0)(2,0) - Where we start - (3,3)(3,3)(3,3) - Our first nice point - (6,6)(6,6)(6,6) - Another helpful point - (11,9)(11,9)(11,9) - One more to show the curve flattening out

STEP 12

Connect these points with a smooth curve, making sure it curves exactly like a square root function should - steeper at the start and gradually becoming less steep.

STEP 13

The graph is a smooth curve starting at point (2,0)(2,0)(2,0), passing through (3,3)(3,3)(3,3), (6,6)(6,6)(6,6), and (11,9)(11,9)(11,9), extending infinitely upward and to the right, but only existing for x≥2x \geq 2x≥2.

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QuestionGraph the function. $f(x)=3 \sqrt{x-2}$
Plot four points on the graph of the function: the le

What is this asking? We need to draw a smooth curve of $f(x)=3\sqrt{x-2}$ by finding some smart points that will help us sketch it properly. Watch out! Remember the domain starts at $x=2$ (not zero) because we can't take the square root of a negative number!

1. Find the domain
2. Find key points
3. Determine shape
4. Plot and connect

Let's think about what values of $x$ we can actually use here.
We've got a square root, and we know we can't take the square root of negative numbers.
So what's inside our square root needs to be positive or zero.
$x-2 \geq 0$

Solving this is super simple - just add 2 to both sides:
$x \geq 2$
This means our graph can only exist for values of $x$ that are 2 or greater.
That's our domain!

Let's start with the easiest point - when $x=2$ .
This is our starting point because it's where the domain begins:
$f(2) = 3\sqrt{2-2} = 3\sqrt{0} = 3 \cdot 0 = \mathbf{0}$
So our first point is $(2,0)$ !

Let's find some more nice points.
When $x=3$ :
$f(3) = 3\sqrt{3-2} = 3\sqrt{1} = 3 \cdot 1 = \mathbf{3}$
We've got $(3,3)$ !

When $x=6$ :
$f(6) = 3\sqrt{6-2} = 3\sqrt{4} = 3 \cdot 2 = \mathbf{6}$
Adding $(6,6)$ to our collection.

One more nice point, let's try $x=11$ :
$f(11) = 3\sqrt{11-2} = 3\sqrt{9} = 3 \cdot 3 = \mathbf{9}$
And we've got $(11,9)$ !

This is a square root function, but it's been modified in two ways: - It's been shifted 2 units right (because of the -2 inside) - It's been stretched vertically by a factor of 3 (because of the 3 outside)

The function will start at our point $(2,0)$ and curve up and to the right, just like a regular square root function, but steeper because of that factor of 3.

Now we can plot our points: - $(2,0)$ - Where we start - $(3,3)$ - Our first nice point - $(6,6)$ - Another helpful point - $(11,9)$ - One more to show the curve flattening out

The graph is a smooth curve starting at point $(2,0)$ , passing through $(3,3)$ , $(6,6)$ , and $(11,9)$ , extending infinitely upward and to the right, but only existing for $x \geq 2$ .