Which equation represents this graph? f(x) = sqrt(x + 2) + 3;

f(x) = - sqrt(x - 2) + 3;

f(x) = sqrt(x - 2) + 3; ,

f(x) = - sqrt(x + 2) + 3

f(x) = - sqrt(x - 2) + 3

Step-by-step explanation:

domain is  

[ 3 , ∞ )

and our range is

( − ∞ , 1 ]

Step-by-step explanation:

Let's look at the parent function:  

√

x

The domain of  

√

x

is from  

0

to  

∞

. It starts at zero because we cannot take a square root of a negative number and be able to graph it.  

√

−

x

gives us  

i

√

x

, which is an imaginary number.

The range of  

√

x

is from  

0

to  

∞

This is the graph of  

√

x

graph{y=sqrt(x)}

So, what is the difference between  

√

x

and  

−

2

⋅

√

x

−

3

+

1

?

Well, let's start with  

√

x

−

3

. The  

−

3

is a horizontal shift, but it is to the right, not the left. So now our domain, instead of from  

[

0

,

∞

)

, is  

[

3

,

∞

)

.

graph{y=sqrt(x-3)}

Let's look at the rest of the equation. What does the  

+

1

do? Well, it shifts our equation up one unit. That doesn't change our domain, which is in the horizontal direction, but it does change our range. Instead of  

[

0

,

∞

)

, our range is now  

[

1

,

∞

)

graph{y=sqrt(x-3)+1}

Now let's see about that  

−

2

. This is actually two components,  

−

1

and  

2

. Let's deal with the  

2

first. Whenever there is a positive value in front of the equation, it is a vertical stretching factor.

That means, instead of having the point  

(

4

,

2

)

, where  

√ 4

equals  2 , now we have  

√ 2

⋅4  equals  2

. So, it changes how our graph looks, but not the domain or the range.

graph{y=2 * sqrt(x-3)+1}

Now we've got that  − 1

to deal with. A negative in the front of the equation means a refection across the  

x -axis. That won't change our domain, but our range goes from  

[ 1 , ∞ )  to  

( − ∞ , 1 ]

graph{y=-2sqrt(x-3)+1}

So, our final domain is  

[ 3 , ∞ )

and our range is  

( − ∞ , 1 ]

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