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Mathematics 7 Online
OpenStudy (javyb13):

PRECALC help will fan and medal and attach the problem

OpenStudy (javyb13):

OpenStudy (anthonyym):

You get the inverse of a function by reflecting it over the line y=x. (You may reacall from algebra that sometimes when want to find the inverse of a function and you switch y and x and solve for y because y=x.)

OpenStudy (javyb13):

okkkk

OpenStudy (anthonyym):

So, let's say you have the point (3,0). It's inverse would be (0,3). You just switch x and y. |dw:1474496889027:dw|

OpenStudy (javyb13):

k i think im kinda getting it

OpenStudy (anthonyym):

Because the inverse of a graph is reflected over the y-axis, the y-axis is the line of symmetry. So the function (graph) and its inverse are symmetrical about the y-axis. (You can elminate the 2nd answer choice).

OpenStudy (javyb13):

ok

OpenStudy (anthonyym):

Then if you draw in or imagine the line y=x there and turn you head, you can see that graph f is definitely not symmetrical with graph h over the line y=x.

OpenStudy (javyb13):

mhm

OpenStudy (javyb13):

ohh ok so then it would be g

OpenStudy (javyb13):

ohh ok

OpenStudy (anthonyym):

OpenStudy (anthonyym):

Ok I confused myself. So the answer is that f and g are inverses symmetrical about the line y=x. (Choice 3)

OpenStudy (anthonyym):

And you can check with the point (3,6), which is on graph f. It's inverse, (6,3), is also on graph g.

OpenStudy (javyb13):

so is that the answer?

OpenStudy (anthonyym):

Yes

OpenStudy (javyb13):

ok thank you so much i think i have another question could you help me with that too?

OpenStudy (anthonyym):

Sure I can try

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