Find the Domain and Range of the following sets of ordered pair:
{(x,y) | y=x^2-2}
{(x,y) | y=√ 4-x}

Question

Find the Domain and Range of the following sets of ordered pair: 
{(x,y) | y=x^2-2}
 {(x,y) | y=√ 4-x}

lgbasallote · Answer

okay...so like i said i'll teach you the 2nd restriction

lgbasallote · Answer

restriction 2 is in form $$y = \sqrt{f(x)}$$

anonymous · Answer

You always get [Math Processing Error] ?

lgbasallote · Answer

let's take d) for example..it has a square root right?

anonymous · Answer

Yep

lgbasallote · Answer

and is there a variable in the square root?

anonymous · Answer

there i

lgbasallote · Answer

therefore it qualifies for restriction 2

anonymous · Answer

uhuh

lgbasallote · Answer

for restriction two you will equate the radicand (expression inside the radical) as greater than or equal to 0

anonymous · Answer

*math nosebleed*

lgbasallote · Answer

for d) $$\sqrt{4-x}$$ you simply do $$4 - x \ge 0$$ that's all it means

lgbasallote · Answer

solve for x

anonymous · Answer

Uhmmm. You always get this lgbasallote
Medals 0

for d) [Math Processing Error] you simply do [Math Processing Error] that's all it means

anonymous · Answer

?

lgbasallote · Answer

refresh

lgbasallote · Answer

it's just an error in the reading of the code i typed which can be easily remedied by refreshing the page

lgbasallote · Answer

just tell me when you're okay

anonymous · Answer

Got it. So X would be $$x \ge -4$$

anonymous · Answer

?

lgbasallote · Answer

4 - x $\ge$ 0
my way is to add x to both sides

4 $\ge$ x

therefore $x \le 4$the domain is x\(\le|) 4 or all values of x less than or equal to 4

lgbasallote · Answer

the domain is $x \le 4$*

anonymous · Answer

Okay. NOw i get it. So how about the one with an exponent? C

anonymous · Answer

Wait. Lol the Range first. for D

lgbasallote · Answer

lol the one with the exponent doesnt matter...if you look at that equation it looks nothing like the erstrictions so the domain is just all real values (which means no restrictions) icannot say the same for its range though coz i cannot compute it mentally

ANYWAY range of d...square both sides to get rid of square root then solve for x

anonymous · Answer

Okay

anonymous · Answer

$$y ^{2} = 4-x$$

anonymous · Answer

X = Y^2 +4 ?

lgbasallote · Answer

good...solve for x

anonymous · Answer

Do I equate i to 0 AGain?

anonymous · Answer

X= -4?

lgbasallote · Answer

oh...no..x = y^2 + 4 is fine...my message was late sent

lgbasallote · Answer

okay look at y^2 + 4

is it a fraction?

anonymous · Answer

NOpe

lgbasallote · Answer

does it have a square root?

anonymous · Answer

Nope. but it's squared

lgbasallote · Answer

nahh we're just concerned IF it is a square root

lgbasallote · Answer

none of the restrictions are satisfied therefore no restirctions...the range is all real numbers

anonymous · Answer

Phew! That was brain wrecking. Okay

lgbasallote · Answer

ikr

lgbasallote · Answer

as for c...i have taught you everything..can you do it yourself?

lgbasallote · Answer

you can ask me if you have doubts on the steps/answer though

anonymous · Answer

Uhmm . The domain is a set of real numbers right?

anonymous · Answer

I still don't get it. I'm a tad confused atm.

lgbasallote · Answer

yep

anonymous · Answer

how do you find the range?

anonymous · Answer

Is it also set of real numbers?

anonymous · Answer

So when it comes to graphing Function what does it mean if a point it circled?

lgbasallote · Answer

forst add 2 to both sides so x^2 will be alone

anonymous · Answer

and then square root both side to get rid of ^2?

lgbasallote · Answer

yep

lgbasallote · Answer

so it's restriction 2

anonymous · Answer

So it's under restriction

lgbasallote · Answer

yes

anonymous · Answer

Then?

anonymous · Answer

I eqaute the radiant to 0?

anonymous · Answer

for restriction two you will equate the radicand (expression inside the radical) as greater than or equal to 0 ?

lgbasallote · Answer

yep

anonymous · Answer

wait. What happens to Y?

lgbasallote · Answer

what do you get as the expression inside the square root?

anonymous · Answer

$$x = \sqrt{2y}$$

lgbasallote · Answer

hmm no...it should be $$\sqrt{y+2}$$ recheck your solution and see where you went wrong

anonymous · Answer

ohh sorry typo

anonymous · Answer

So it plus. So  x = $$\sqrt{y+2}$$

anonymous · Answer

So do I equate y+ 2 to 0?

lgbasallote · Answer

y+2 $\ge$ 0

anonymous · Answer

Ohhhh. So that' how it goes but how will you know if it's greater than or less than?

lgbasallote · Answer

it's always greater than or equal to 0 i mean at first

lgbasallote · Answer

if y or x becomes negative then you switch the sign

anonymous · Answer

Ohhh. Okay now I get it. Thank!

anonymous · Answer

Now on for the next one. Do you know I have  12 more question to go. *sighs* and it's all due on june 8. Guess I'll be pulling an all nighter.

lgbasallote · Answer

well im not o.O sorry lol..pretty sure lots of others here can answer them though

anonymous · Answer

Yep. thanks for you help! and the medal ^^

lgbasallote · Answer

you were a good student :p

anonymous · Answer

:))) You're a good teacher xP