# intersection of two graph

i would like to clarify some questions from GRE,which at first seems a little difficult to understand,suppose that we have some function

$f(x)=|2*x|+4$ and graph of this function is given

where is following question: For which of the following functions g defined for all numbers x does the graph of g intersect the graph of f ? and 4 possible answers are

A.g(x)=x-2
B.g(x)=x+3
C.g(x)=2*x-2
D.g(x)=2*x+3
E.g(x)=3*x-2


first what i did not understand what does mean

For which of the following functions g defined for all numbers x does the graph of g intersect the graph of f ? does it means that which g intersect of f for all x or?answer is E,i have guessed that for intersection we should have $|2*x|+4=g(x)$,clearly C and D no,because they are parallel lines with common slopes,so whe should have

$|2*x|+4=x-2$

or $|2*x|+4=x+3$

or $|2*x|+4=3*x-2$

after some calculation we will get $|2*x|=x-6$

or $|2*x|=x-3$

|2*x|=3*x-6

first can't be ,because $|2*x|>=x$ second also using the same rule,only one left

$|2*x|=3*x-6$ by using solving,for example if $x>0$ $2*x=3*x-6$ $x=6$

if $x<0$

$-2*x=3*x-6$

$x=6/5$

is there any short way to solve it?

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I would just think of it geometrically. Look at the $y$-intercept and slope of one of the lines (the graph of $g$ will intersect the graph of $f$ if $g(0)<4$ and the slope of $y=g(x)$ is greater than $2$ or less than $-2$, for example). It seems you used similar reasoning for some parts; but you can do it for all of them. You don't want to waste time doing needless calculation... –  David Mitra Jun 20 '13 at 14:06
this criteria is because of modulus operator? –  dato datuashvili Jun 20 '13 at 14:09
If I understand you correctly, yes. You have to consider "both sides" of the graph of $f$. For example, in E., the graph of $g$ will intersect the graph of $f$ at some $x>0$, since $g(0)<f(0)$ and the slope of $y=g(x)$ is greater than the slope of the "right side of $f$". (Sorry for the sloppy phrasing). If there were a part F.: $g(x)=-3x-2$, then the graphs wold intersect for some $x<0$. Perhaps sketching the graphs would help... –  David Mitra Jun 20 '13 at 14:14
ok thanks very much,i think that during the GRE test there will not be required too much calculation –  dato datuashvili Jun 20 '13 at 14:17
To put it simply, the graph of $f$ is a V shape, and each of the possible answers is a line. Choices A and B would look like $\underline{V}$, just with different amounts of space between the V and the underline, and choices C and D would look like V /, again just with different spacing between the V and the /. Only in choice E will the slash cut through the vee. –  cobaltduck Jun 20 '13 at 14:29

You only need the two graphs to intersect at one point. If this is so for a single value of $x$, then the corresponding item will be a solution.
You are correct that "E" is the only answer. A quick way to solve the problem is to note that the graphs of all the items are straight lines; so you can deduce things from geometric reasoning. Just look at the $y$-intercepts of the lines and compare the slopes of the lines to the slopes of the line segments comprising the graph of $f$. It would also be helpful, for each item, to separate the problem into two, natural, parts: Does the given line intersect the graph of $f$ at some $x\ge0$? Does the given line intersect the graph of $f$ at some $x<0$?
E: The $y$ coordinate of the $y$-intercept of the line here is $-2$, which is less than $f(0)=4$, and the slope is $3$. Since the slope of the line segment given by $y=f(x), x\ge 0$ is $2<3$, the graph of $y=g(x)$ will intersect the graph of $y=f(x)$ for some $x>0$ (over $[0,\infty)$, the graph of $g$ rises more quickly than the graph of $f$).