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 !!!
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 !!!
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
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
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!!!
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.