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    Hi just wondering why does a ii) only have one solution? I thought both lines meet at 2,0 and -2,0... Name:  1456385988605646875276.jpg
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    Draw the line y = -x on the graph, you should see that they only cross once.
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    (Original post by coconut64)
    Hi just wondering why does a ii) only have one solution? I thought both lines meet at 2,0 and -2,0... Name:  1456385988605646875276.jpg
Views: 90
Size:  317.2 KB thanks.
    It will only meet the graph once.

    Draw the line y=-x and you'll see it'll intersect only once; if you're struggling to visualise it, a table of values is always the best thing to do
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    This is a question from Solomon Papers, I think.
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    (Original post by coconut64)
    Hi just wondering why does a ii) only have one solution? I thought both lines meet at 2,0 and -2,0... Name:  1456385988605646875276.jpg
Views: 90
Size:  317.2 KB thanks.
    Look at the gradient of the f(x) graph in the negative region, it goes from -2 to 0 in the x-direction and that makes it go from 0 to 2 in the y-direction, so it has a gradient of -1.

    That's the same gradient as the line y=-x. so if you plot y=-x, it will always be two units behind (x-direction) wise f(x) in the negative region since they have the same gradient and hence increase/decrease at the same rate, but their 'starting point' is different.

    Edit to add: in a more formal manner, you can define f piecewise such that it is y=-x+2 for x \leq 0, then in the left-half plane, it will intersect y=-x if and only if -x = -x + 2 \iff 2=0. Which is false. So it won't intersect y=-x in the left-half plane.
 
 
 
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