The line represented by which of the following equation does

@EvaJager/Bunuel: How did we conclude that the 2 parabolas (one that is shifted up vertically w.r.t the other) does NOT intersect each other ?
I guess i am missing something basic here.
Thanks!

For the same value of x, we get some y for one expression and y + 1 for the other expression. y cannot be equal to y + 1.

The other way to solve this question is to create a graph for -2,-1,0,1,2.

Now put these values in the option to see which option doesn't intersect.
This solution is not meant for those who are aware of parabola & base shift or twist

Hope it helps

Manju,

concept of slope for lines & parabolas are different. Bunuel, please correct if I am wrong. Also please help to solve this problem if its a GMAT type question.

The general form of parabolic equ. is y^2= 4ax which implies the axis is x or x^2 = 4ay where axis is y.
We have a similar form as x^2 = 4ay.
here the vertex is origin.

So if we have same values of x and y but constant term changes then we will have parallel parabolas.
This is same as for straight line which are parallel for different values of constant term c
ax + by +c1 = 0 and ax +by+ c2 =0

We have quadratic equations. These equations when drawn give parabolas, not lines. The question is: which of the following parabolas does not intersect with the parabola represented by y=3x^2+5x+1.

This CANNOT be transformed to the question: "which of the following parabolas is parallel to the parabola represented by y=3x^2+5x+1." In the wast majority of cases the word "parallel" is used for lines. Well, we can say that concentric circles are parallel, BUT GMAT, as far as I know, uses this word ONLY about the lines. Next, the word "parallel" when used for curves (lines, ...) means that these curves remain a constant distance apart. So strictly speaking two parabolas to be parallel they need not only not to intersect but also to remain constant distance apart. In this case, I must say that this cannot happen. If a curve is parallel (as we defined) to the parabola it won't be quadratic: so curve parallel to a parabola is not a parabola.

Hi all,

Now we see from the statement that y = 3x^2+5x+1 is a parabola.

The y intercept represents the vertex therefore if +1 is replaced by +2 such as in answer choice C the parabola only move upwards but means that it will never intersect with the original equation.

Therefore answer is C

Hope this helps
Cheers
J

The line represented by which of the following equation does not intersect with the line represented by y = 3x^2+5x+1

Calculate Discriminant (D) for each equation :\(\sqrt{b^2-4ac}\)

y = 3x^2+5x+1 ==> \(\sqrt{13}\) -- cutting Y axis at 1 -- to calculate intercept put x=0
A. y = 2x^2+5x+1 ==> \(\sqrt{17}\) -- D > \(\sqrt{13}\) means curve is below original curve cutting Y axis at 1 -- cutting at same point.
B. y = x^2+5x+2 ==> \(\sqrt{17}\) -- D > \(\sqrt{13}\) means curve is below original curve and Y intercept at 2-- cut is unavoidable.
C. y = 3x^2+5x+2 ==> \(\sqrt{1}\) -- D < \(\sqrt{13}\) means closest to X axis -- cutting y axis at 2 above 1 -- cutting right above on Y axis and curve is also passing above as D = 1.
D. y = 3x^2+7x+2 ==> \(\sqrt{25}\) -- D > \(\sqrt{13}\) means curve is below original curve and Y intercept at 2-- cut is unavoidable.-- not plotted on attached graph.
E. y = x^2+7x+1 ==> \(\sqrt{45}\) -- D > \(\sqrt{13}\) means curve is below original curve cutting Y axis at 1 -- cutting at same point.

Refer following graph to relate the nature of equations and value of D.

I'd emphasize that none of the equations in this question represent lines, despite what the question appears to say. The equations represent parabolas, and you don't need to know about parabolas for the GMAT.

There is one concept in this question that is occasionally tested - the concept of translation. If you have any equation at all in coordinate geometry, say:

y = x^2

that will be some curve in the coordinate plane (technically it will be a 'parabola', or U-shape). If you then modify the equation by adding a constant on the right side, say by adding 5:

y = x^2 + 5

then the graph of this new equation will look exactly the same as the graph of the first equation, except that it will be exactly 5 units higher. So when we add a constant on the right side of an equation, we're simply moving the picture of the equation up or down.

I think Bunuel and Ian have provided sufficient information to solve this problem.

Lines are always represented by LINEAR equations (equations that have maximum degree of the variables as 1). A quadratic equation (max. degree =2) can NEVER represent lines.

I would like to add one thing that people who are not familiar with 'conics', usually do not remember that \(y^2=4ax\) is the standard equation of a parabola. One way to eliminate such rote learning is to look at an equation and plot 2-3 points and see what shape of the curve do you get and then proceed from there. GMAT does not require you to remember fancy names.

\(y^2=4ax\) and \(y^2=4ax+Z\), where Z is any value (4,5,7.8,0.4 etc). These curves belong to 'same family' of parabolas with the only exception that these 2 curves will have their vertices offset by 'Z' amount.

Finally, for the curious minds out there, the attached picture shows all the possible combinations of 'simple' parabolas.