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Algebra 26 Online
OpenStudy (lexy148):

How do you graph and solve for y= square root x-2 if the x points are 0,-1,-2,1,2

jimthompson5910 (jim_thompson5910):

is the equation this? \[\Large y = \sqrt{x-2}\] or this?\[\Large y = \sqrt{x}-2\]

OpenStudy (lexy148):

the 1st one

OpenStudy (lexy148):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

if we replace x with 0, what result do you get \[\Large y = \sqrt{x-2}\] \[\Large y = \sqrt{0-2}\] \[\Large y = ???\]

OpenStudy (lexy148):

-2 ?

jimthompson5910 (jim_thompson5910):

you'll have \[\Large \sqrt{-2}\] BUT you cannot take the square root of a negative number. The result isn't a real number. So the calculator should say NAN or undefined or something

jimthompson5910 (jim_thompson5910):

did you type in `sqrt(0-2)` or `sqrt(-2)` into your calculator?

OpenStudy (lexy148):

yeah thats what it said when i did it in the calculator

jimthompson5910 (jim_thompson5910):

so because of this error, we don't include x = 0 in the domain

OpenStudy (lexy148):

i typed sqrt -2 not 0-2

jimthompson5910 (jim_thompson5910):

which calculator are you using?

OpenStudy (lexy148):

we wouldnt include any negative numbers in sqrt then

jimthompson5910 (jim_thompson5910):

yes you cannot have negative numbers under the square root eg: \(\Large \sqrt{-1}\) is not allowed

OpenStudy (lexy148):

a scientific calculator

jimthompson5910 (jim_thompson5910):

ok, what happens when you replace x with -1 instead of 0? what result do you get?

OpenStudy (lexy148):

ok so i wot put any negatives then thx

OpenStudy (lexy148):

i get -3

jimthompson5910 (jim_thompson5910):

the goal is to have x-2 never be negative so we can force it to be 0 or positive, ie force it to be greater than or equal to 0 \[\Large x-2\ge 0\] solve for x to get \[\Large x\ge 2\] so you are allowed to input values of x that are either 2 or larger than 2

jimthompson5910 (jim_thompson5910):

you'll find that \[\Large \sqrt{-3}\] isn't a real number. So x = -1 isn't allowed to be an input either

OpenStudy (lexy148):

ok i see so only numbers 2 or more, no negatives

jimthompson5910 (jim_thompson5910):

yep

jimthompson5910 (jim_thompson5910):

if x = 2, then what is the value of y?

OpenStudy (lexy148):

ok thank you and how would you sqrt 0 since 2-2 is 0

jimthompson5910 (jim_thompson5910):

the square root of 0 is 0

jimthompson5910 (jim_thompson5910):

your calculator should be able to calculate that if not, then use this calculator http://web2.0calc.com/

OpenStudy (lexy148):

nevermind i got it

jimthompson5910 (jim_thompson5910):

ok one more: if x = 6, then what is the value of y?

OpenStudy (lexy148):

thank you for helping me :)

OpenStudy (lexy148):

it would be 4

jimthompson5910 (jim_thompson5910):

the `x-2` part is 4 when x = 6 but I want the value of y \[\Large y = \sqrt{x-2}\] \[\Large y = \sqrt{6-2}\] \[\Large y = \sqrt{4}\] \[\Large y = ??\]

OpenStudy (lexy148):

oh srry it would be 2

jimthompson5910 (jim_thompson5910):

correct, so when x = 6, the value of y is y = 2 that means we have the point (6,2) earlier you found that when x = 2, the value of y was y = 0 so the point (2,0) You make a table of x,y values and plot each ordered pair as a point on the xy grid. Once you have enough points, you can draw a curve through them all to complete the graph

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