Which of the following is a possible equation for the above graph?

Question

https://gmatclub.com/forum/download/file.php?id=27751 Which of the following is a possible equation for the above graph? A) x^3 B) x^3 -1 C) 3x^3 + 2x D) 3x^3 - 2x E) x^3 + 3x^2 - x + 2 https://i30.tinypic.com/2njj31c.jpg Graph.png

learner · Accepted Answer

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.

MohitRulz · Answer

are we goin to use equation of line y=mx+b here?

snipertrader · Answer

D it is
I don't think we can use the slope formula here... its a simple case of solving for X or Y and seeing the corresponding points on the graph.

What the source?

chunsli · Answer

Just use elimination, or substitution, when u sub in (0,0), u eliminate two choices, and when u sub in a small number, u elminate A and C. then u are left with D

MohitRulz · Answer

thank u 'Learner' for writing so much.Thanks a lot.(deserver a kudos)
Sniper, source is Princeton prep course.

amit2k9 · Answer

well from the graph one can see three solutions - (0,0),(x,0) and (-x,0).

substituting for y = f(x) we have,

C and D options left.

With a positive value for X, Y = 0 as can be seen in the positive half of the graph.

D is the only option left.

bschool3 · Answer

Pretty sure calculus is not in the scope of the GMAT, so I apologize if this is a waste of time, but differentiation comes in handy here. There should be 2 points where the first derivative equals zero, i.e. a quadratic equation with 2 distinct roots for the local minima and maxima. This rules out A,B,and C. To decide between D and E, we apply second order conditions;we know that there is one inflection point at the origin, therefore the second derivative must = 0 where x = 0. For D, we have dy/dx = 9x^2 - 2; d2y/dx2 = 18x = 0, gives x=0, as required. Choose D. For E, we have dy/dx = 3x^2+6x-1. d2y/dx2 = 6x+6 =0. x is not 0. The answer is D

hdwnkr · Answer

No need to know the equation of this scary graph.

According to the graph, for x = 0. y = 0. This is possible in only three options. A,C and D

For x^3, when x is positive, x^3 would never come below the x axis and give a negative value. Hence, A is discarded.

From the other two options, take x as 0.5 and 1. Option D gives the negative dip.

GMATisLovE · Answer

The answer is D as follows.

The graph shows that at x=0, y=0. Putting the value of x=0 in the above 5 equation will give that only A, C and D are left. B and E are out.
Now we will have to calculate the slope. So using the concept of differentiation the formula for slope are as follows.
A -> 3 x^2 
C -> 9 x^2  + 2
D -> 9 x^2  - 2

From the graph we also know that the slope at x=0 is -ve
Putting the value of x=0 in the above 3 equations, we get

A -> 3 0^2  --> 0
C -> 9 0^2  + 2 --> 2
D -> 9 0^2  - 2 --> -2

Only D satisfied the required condition,

Hence answer is D

hellosanthosh2k2 · Answer

A) x^3
B) x^3 -1
C) 3x^3 + 2x
D) 3x^3 - 2x
E) x^3 + 3x^2 - x + 2

one quick approach, i can think of,

First looking at the graph, there are up and downs, it means, there is no correlation between x and y.
With this idea, let us attack the answer choices.

1. y = x^3 , when x is going to increase, y is also going to increase, there is co-relation, so this can't be the equation
2.  y = x^3 -1 , when x is going to increase, y is also going to increase, there is co-relation, so this can't be the equation.
3.  y = 3x^3 + 2x , when x is positive, y is also positive, when x is negative, y is also negative, no chance of up and down, so this can't be the equation.

So left which D and E.

We can eliminate one of them, by looking at graph, if x = 0, then y = 0, only choice D satisfies.

That should be our answer (D)

GMATGuruNY · Answer

Since the graph passes through (0, 0), the correct answer must yield y=0 when x=0.
Eliminate B and E, which do NOT yield y=0 when x=0.

The graph has three x-intercepts.
Eliminate any remaining answer that will not yield three x-intercepts.

Since A will yield an x-intercept only at (0, 0), eliminate A.

Calculate the x-intercepts for C:
 0 = 3x^3 + 2x 
 0 = x(3x^2 + 2) 
 x=0  or  x^2 = -\frac{2}{3} 
Since  x^2 = -\frac{2}{3}  has no real solutions, C will yield an x-intercept only at (0, 0).
Eliminate C.

D

Crytiocanalyst · Answer

The key to solving the same is eleminating each option in accordance with graph:
(0,0) is a point in the graph and another negative value of x y tends to be zero

A) x^3
(0,0) is satisfied however we are not able to determine the other point

B) x^3 -1
at x=0 y not equal to zero hence eleminate

C) 3x^3 + 2x
(0,0) is satisfoed x(3x^2+2)=0 provides complex roots which Gmat doesn't take into account hence out

D) 3x^3 - 2x
(0,0) is satisfoed and a real root is possible

E) x^3 + 3x^2 - x + 2
(0,0) is not satisfoed 
Hence IMO D

arushi118 · Answer

We have 3 roots, with one of them at x=0.
Only A, C, D have a root at x=0.
A has only x=0 as root - so not A
Between C and D:
C = x(3x^2 + 2) -> when equated to 0, it will give x^2 = a negative value - so not correct
D: This will give 3 roots when equated to 0. - One at x=0, and the other at a positive and a negative value of x - exactly as the graph shows.

Which of the following is a possible equation for the above graph?

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Which of the following is a possible equation for the above graph?

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