In the xy-plane, a parabola intersects with axis-y at point

Question

In the xy-plane, a parabola intersects with axis-y at point (0,y). Is y < 0 ?

(1) The vertex of parabola is (2,-5)
(2) The parabola intersects with axis-x at point (-2,0) and (6,0)

Bunuel · Accepted Answer

Though it's possible to solve this question algebraically the easiest way will be to visualize it and draw on a paper. (1) The vertex of parabola is (2,-5) --> the vertex is in the IV quadrant: if the parabola is downward it'll have negative y-intercept, but if it's upward then it can have positive as well as negative y-intercept. Not sufficient. (2) The parabola intersects with axis-x at point (-2,0) and (6,0) --> now if the vertex is above x-axis then parabola will have positive y-intercept and if its vertex is below x-axis it'll have negative y-intercept. Not sufficient. (1)+(2) As from (1) the vertex is below x-axis then from (2) we'll have that parabola must have negative y-intercept. Sufficient. You can look at the diagram below to see that a parabola passing through the given three points must have negative y-intercept only. MSP139819db6ebe95a9e2a900005889a632a09g5628.gif Answer: C. Hope it's clear.

arjunbt · Answer

Thanks Bunuel for the explanation,
I was thinking since option B gives us (-2,0) and (6,0) vertices, with this we can assume that the parabola is directed upwards and it's Y intersect will be -ve and hence sufficient, but
after reading your explanation I understand it better

1+ Kudos to you.

Bunuel · Answer

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Coordinate Geometry: math-coordinate-geometry-87652.html

All DS Coordinate Geometry Problems to practice: search.php?search_id=tag&tag_id=41
All PS Coordinate Geometry Problems to practice: search.php?search_id=tag&tag_id=62

sr2013 · Answer

Hi Bunuel

What is wrong in the following approach

1) Insufficient 
 The vertex of parabola is (2,-5) --> the vertex is in the IV quadrant: if the parabola is downward it'll have negative y-intercept, but if it's upward then it can have positive as well as negative y-intercept. Not sufficient

I was thinking from option 2 we can find the product of the roots and sum of the roots 
product of roots = c/a = -12

and sum of roots = -b/a = 4

hence come up with an equation y = x^2-4x-12

and from the question stem we have that it intercepts at 0,y

Putting x = 0 we get y = -12 (negative) and B alone is sufficient .

Bunuel · Answer

How did you get y=x^2-4x-12 from c/a=-12 and -b/a=4? You cannot solve c/a=-12 and -b/a=4 to get unique values of a, b, and c. For example if a=2, c=-12, and b=-4 you'll get 2x^2-24x-8=0: MSP2141ga1ih773ii7e3di00002h2000h2fiibia6c.gif If a=-1, c=12, and b=4 you'll get -x^2+4x+12=0: MSP24971d0ia65iig79i4gg000037eb4cg662i7adih.gif Or in other words infinitely many parabolas have x intercepts at -2 and 6. You cannot get unique equation only from that info.

kd1989 · Answer

Hi Bunuel

I have tried solving it using standard expressions for parabola
For parabola y= ax2+ bx+ c, standard vertex is located at point (-\frac{b}{2a}, c-\frac{b^2}{4a}).

From a) we know the value of vertex as (2,-5) 
by putting the value in standard vertex we can get c=-4
it is also given in the question stem that parabola intersects y axis at (0,y)
from this we can get the value of y as -4. which is sufficient to answer the question.

Please let me know whats wrong with this approach.

Bunuel · Answer

The post you are quoting has an answer to your question. You cannot solve for c. P.S. Please read this: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum)

chetan2u · Answer

A parabola has a vertex and a similar curve on both sides..
the vertex can be MIN or MAX value depending on the way parabola opens up...

1. The vertex of the parabola is (2,-5) 
 we know vertex lies below the x-axis , but --
a) if it opens upwards- the intersect with y-axis can be either below x-axis or above it..
b) if it opens downwards- the intersect with y-axis will be below x-axis ..
Insuff

2. The parabola intersects with axis X at point (-2,0) and (6,0) 
 we know vertex would lie at (-2+6)/2 = 2 as x, BUT we cannot determine if VERTEX is above or below x-axis..
whereever vertex lies, the intersect will lie on that point .. 
since the curve moves from 2 to -2 in that Quadrant -

a) if it opens upwards- the intersect with y-axis will be below x axis, as the vertex will be below x-axis..
b) if it opens downwards- the intersect with y-axis will be above x-axis, as the vertex will be above x-axis ..

combined-
statement I tells us that the vertex is below the x-axis and statement II tells us that if vertex is below x- axis the intersect is also below x-axis..
so y<0..
Suff
C

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

In the xy-plane, a parabola intersects with axis-y at point

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

In the xy-plane, a parabola intersects with axis-y at point

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards