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Re: Equation of the graph? [#permalink]
10 Aug 2009, 21:00
This post received KUDOS
The graph is a plot of y=f(x) ,which we have to find. Look at the graph closely. The graph cuts at (0,0)
So when x=0, y =0 . This eliminates option B and E. When you substitute 0 in option B, y= x^3 - 1, if x=0 -> y=0^3 -1 =-1. The corresponding co-ordinate is 0,-1 which is not the case with the graph
When you substitute 0 in option E , y= x^3+3x^2-x+2 , if x=0 y= 2. Again, the corresponding co-ordinate is 0,2 which is not the case.
Now we are left with options A,C and D.
In A, y =x^3. If x>0, y should be greater than 0. But this is not the case in the given graph. The graph has points in 4th quadrant which is (x,-y). So option A can be ruled out.
Now consider C , y=3x^3 + 2x. Again if x>0 , y should be greater than 0. Ex. if x=1 , y= 5. if x=1/10 , y = 0.003+ 0.2 = 0.203. But this is not the case in the given graph. The graph has points in 4th quadrant which is (x,-y). So option C can be ruled out.
Now consider D, y= 3x^3 -2x , In this case , for x>0 , y can be gretaer than 0 or less than 0. For ex, if x=1/10, y= 0.003-.02= -0.017. If x=1, y =1. For x=2, y=22. So for x> 0, Y can be less than or greater than 0, spanning I and IV quadrant. Therefore option D is correct.