graphing: find point on graph, find eqn of y shifted 6 units

[pendulum]

y=sqrt(x-2)+5. Which point lies on the graph?

a. 7,6

b. 0,5

c. -2,5

d. 3,6

I forgot my book at school, I think its C but I dont know how to get there. Please help.

One more question: graph y=sqrt(x) is shifted 6 units down and 1 unit right. What equation represents the new graph?
a. y=sqrt(x+1)-6
b.y=sqrt(x-6)-1
c.y=sqrt(x-1)-6
d.y=sqrt(x+6)-1

I think it is A because 1 unit right is positive and 6 units down is negative but I am not so sure.

 

[Deleted member 4993]

Re: graphing

y=sqrt(x-2)+5. Which point lies on the graph?

a. 7,6

b. 0,5

c. -2,5

d. 3,6

I forgot my book at school, I think its C but I dont know how to get there. Please help.

Put those (x,y) in the given equation and see which one satisfies. For example) (5,0)

0 = sqrt(5-2) + 5 <<< does not satisfy - so (5,0) is not on the graph edited

One more question: graph y=sqrt(x) is shifted 6 units down and 1 unit right. What equation represents the new graph?
a. y=sqrt(x+1)-6
b.y=sqrt(x-6)-1
c.y=sqrt(x-1)-6
d.y=sqrt(x+6)-1

y = f(x-h) + k

is y = f(x) with 'k' units shifted up and 'h' units shifted to left.

I think it is A because 1 unit right is positive and 6 units down is negative but I am not so sure.

 

[pendulum]

Re: graphing

Thank you for your help.

For #1 you put 0 as y..isn't it x though?

Because when i plugged in 7 for y and 6 for x it worked

But when I plugged in 6 for y and 3 for x it worked as well.

I also used the y=f(x-h)+k
I got y=sqrtx-1 -6

Do i switch -1 to +1 and -6 to +6?

 

[pendulum]

Re: graphing

nvm

 

[Deleted member 4993]

Re: graphing

Thank you for your help.

For #1 you put 0 as y..isn't it x though?<<< You are absolutely correct - I fixed it above

Because when i plugged in 7 for y and 6 for x it worked

But when I plugged in 6 for y and 3 for x it worked as well.

I also used the y=f(x-h)+k
I got y=sqrtx-1 -6

Do i switch -1 to +1 and -6 to +6?