BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

How to show that a curve lies above the x-axis?

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Dec 09, 2012 9:11 am
jspake wrote:Hello,
I couldn't figure out how to solve the question below:

The equation of a curve is y = x^2 − 3x + 4.
Show that the whole of the curve lies above the x-axis.

Please help..
Doesn't sound like a GMAT question to me.
What are the 5 answer choices?

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
User avatar
GMAT Instructor
Posts: 613
Joined: Thu Mar 22, 2007 6:17 am
Location: madrid
Thanked: 171 times
Followed by:64 members
GMAT Score:790

by kevincanspain » Sun Dec 09, 2012 12:35 pm
Remember the quadratic formula ?

E = ax^2 + bx + c = 0 iff x= (-b +- sqrt(b^2 - 4ac))/2a


if b^2 - 4ac < 0, E=0 has no real roots and E and a have the same sign for all real numbers x
Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sun Dec 09, 2012 12:46 pm
jspake wrote:Hello,
I couldn't figure out how to solve the question below:

The equation of a curve is y = x^2 − 3x + 4.
Show that the whole of the curve lies above the x-axis.

Please help..
Since we're helping you with your homework (Algebra II perhaps?), here's another approach.

Use the Complete the Square technique.
y = x^2 − 3x + 4
y = x^2 − 3x + 2.25 - 2.25 + 4 (completed the square)
y = (x - 1.5)^2 - 2.25 + 4 (factored)
y = (x - 1.5)^2 + 1.75 (simplified)

So, this parabola has its vertex at coordinates (1.5, 1.75), which means the vertex is above the x-axis.
Since the parabola opens up (which I'll leave you to convince your teacher of :-) ), this parabola must lie above the x-axis.

Aside: For those studying for the GMAT (i.e., the vast majority of viewers), the Completing the Square technique is out of scope for the GMAT.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Sun Oct 28, 2012 8:09 pm
Thanked: 1 times
Followed by:1 members

by savvysatyam » Mon Dec 10, 2012 2:03 am
Since a(the coefficient of quadratic equation) is greater than 0 and the determinant(b2)< 4ac, therfore it will be a parabola with face upwards and since determinant is not equal t zero it will be above x-axis and not touch it
Top
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Sun Dec 09, 2012 6:06 am

by jspake » Thu Dec 13, 2012 1:39 am
Thanks a lot guys.. :)
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 194
Joined: Mon Oct 15, 2012 7:14 pm
Location: India
Thanked: 47 times
Followed by:6 members

by The Iceman » Thu Dec 13, 2012 10:39 pm
Suppose we have a quadratic ax^2+bx+c=0

First, the sign of the coefficient of x^2 determines whether the parabola opens upwards or downwards. (a>0=> parabola opens up, and a<0=> parabola opens downwards)

Case 1: D=0 => Touches x axis at one point (significance: both roots real and equal)

Case 2: D>0 => Cuts x axis at two point (significance: the two roots are real and distinct)

Case 3: D<0 => Does not touch/cut/intercept x axis at any point (significance: both are non real roots)
Top
Post new topic Post Reply
• Page 1 of 1