
Use the graph that shows the solution to f(x)=g(x) .
f(x)=1/x−2
g(x)=x−2
What is the solution to f(x)=g(x) ?
Select each correct answer.
−1
1
2
3
would I plug in those vales for x? I am unsure what to do
Yep... ...I'd just check the intersection points
On the top graph the points appear to be ( 1 , -1) and (-1, - 3)
So putting x = 1 into both functions results in y = -1
So this is correct
And putting -1 into each function results in y = -3
So....this is also correct
So....we don't even need to check the bottom graph...the top graph has to be the correct one !!!
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The question actually isnt regrading the graphs its asking
What is the solution to f(x)=g(x) ?
Select each correct answer.
−1
1
2
3
it either -1,1,2,or 3 for the answer
OK....I guess they are asking for the x values where the graphs intersect
These are
x = 1 and x = -1
Sorry I missed that !!!
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well wouldnt we put them to equal? and that would lead to x=3 and x=1 , so the answer would 3 and 1 ?
Jenny....It looks like they just want you to select the x values of their intersections
These would be
-1
1
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That is kinda' what we did...!!!
If we put x = 1 into 1/x - 2 we get -1
If we put x = 1 into x - 2 we get -1
So ....the same x value plugged into both functions gives the same y value in both functions ....so (1, -1) is an intersection point
Check that x = -1 plugged into both functions gives y = -3
So....the same x value in both functions gives the same y value in both functions....so ( -1 , -3) is the other intersection point
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oh wait! I see now that makes sense! Im sorry i was thinking something totally different yes that would be the correct way . I am so sorry I was just trying to make sense of it all . Thank you for going over it more with me and answering my questions. ![]()
HAHAHA!!!.....this stuff isn't easy.....I hope I make it more understandable
BTW....we could set the functions equal and solve for x
1/x - 2 = x - 2 multiply through by x
1 - 2x = x^2 - 2x simplify
1 = x^2
x^2 = 1 take both roots
x = ±√1
So x = 1 or x = -1
This is sometimes faster than trying to look at a graph!!!
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You do very well all the time. Your explaintations really help. And oh yes, I am more familair with that then looking at the grah but it is good to know both ways. ![]()