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Problem on quadratic polynomial.

  1. Dec 17, 2011 #1
    1. The problem statement, all variables and given/known data

    The graph of the quadratic polynomial , y=ax2+bx+c is as shown below in the figure :

    http://postimage.org/image/nvkxv74yd/ [Broken]

    Then :

    (A) b2-4ac<0
    (B) c<0
    (C) a<0
    (D) b<0

    2. Relevant equations

    If y=0 , then ax2+bx+c=0
    Then ,
    x = {-b+-sqrt(b^2-4ac)}/2a

    3. The attempt at a solution

    Hmm tally image please...
    When parabola is intersecting x axis then y=0 and
    Case 1 : x<0
    Case 2 : x>0
    if x >0 or x<0
    then b2-4ac >0
    So b2 > 4ac

    Now what ...:confused:

    Please help !

    Thanks in advance. :smile:
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 17, 2011 #2

    Tom Mattson

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    You're thinking too hard. Think about the concavity of the graph (that is, the direction in which the parabola opens).
  4. Dec 17, 2011 #3
    Thanks for a hint ex-mentor !
    Parabola is opening in the -y axis direction , right ?
  5. Dec 17, 2011 #4
    If A was correct:
    it would mean that y=ax^2+bx+c =0 would have no real solution, which would mean that the y(x) would never be zero, which is wrong by the graph.

    If B was correct:
    it would mean that for x=0 the y(x)=y(0) would take the value c <0 which by the graph is incorrect. y(0)>0.

    Now C :
    You check y(x) derivative.
    y'(x)= 2ax+b.
    and from that we see that y'(0)= b.
    Though since your MAXIMUM is at x=0, it means that y'(x) must be zero at that point y'(0)=0.

    Checking on D.
    The way to see it is to look at its second derivative:
    y''(x)= 2a.
    you see from the graph that a has to be a<0 because of the concavity.
  6. Dec 17, 2011 #5
    Please don't go so sophisticated. We haven't been taught limits and derivatives and integrals till now because I am after all in class 10th , 15 years. However your answer is correct and that was only given in answer key. Can it be done without involving introductory calculus ?

    You are a new member. Please don't give full answer as its against forum rules. :redface:
    Just alerting you.
  7. Dec 17, 2011 #6
    Ups sorry...I didn't know...I will keep that in mind.

    Oh well at least for the 2 firsts i think you dont need calculus.
    Now the rests, you can check anytime.
    Try drawing for example:

    with a,b,c so that you will have your cases.

    eg for case A it is y=x^2 + x + 1 b^2-4ac= -3 <0

    and see how it works :) then see which one fits better
  8. Dec 17, 2011 #7

    I like Serena

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    Hi sankalpmittal! :smile:

    Let's try and categorize your drawing of the parabola a bit.

    How many roots do you have?
    And what consequence does that have for your solution formula?

    At which x coordinate do you have the top of the parabola?
    And at which side of the x-axis is the top?
    Can you substitute that in the generic parabola equation and draw some conclusions?

    You already saw it opens in the negative y-direction.
    What happens if you substitute a very large x-value in the parabola equation?
    What would the sign be of the resulting y-value?
  9. Dec 17, 2011 #8

    Tom Mattson

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    Yes that's right. What, if anything, does that imply for the coefficients a, b, or c?
  10. Dec 17, 2011 #9


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    When x is very large, y is approx equal to ax2
    Since x2 is going to be positive, the sign of y in this region tells you something about a.
  11. Dec 18, 2011 #10
    Many thanks to NascentOxygen , Tom Mattson , ILS and Morgoth for their hints ! :smile:

    I think I figured out this problem.

    The first option A i.e. b2-4ac<0 is incorrect because that parabola is yielding real roots. Option A is out.

    When that parabola is cutting the y axis then of course x=0
    this implies that

    As we know from graph that y is positive so c is also positive. This implies that c>0.
    But option B says that c<0 which is wrong. Option B is out.

    Now when parabola is intersecting on first quadrant (+) of x axis , then y=0 (we can also take -x axis also.)
    As we know that
    m = Δy/Δx
    We know from above equation
    m = (y-c)/x where c is y intercept. As y = 0 then also c =0 because Δy=0 here. We aren't taking in account y intercept because its 0 !

    This implies y=c=0

    So we know b2-4ac>0 as its yielding real roots.
    so b2-4.a.0>0

    But option D says that b<0 which is incorrect as we found that b>0 so option D is out.

    Now only option which is left is C which has to be correct because ,
    x = {-b+-sqrt(b^2-4ac)}/2a
    as c=0

    I get 2 values :

    x= (-b+b)/2a

    x= -2b/2a
    Now here x is positive because axis is +x. To make "x" positive "a" has to be negative as we know before that "b" is positive. This implies that a<0.

    I got the correct answer but am I methodically correct ?

    Thanks again !:smile:
  12. Dec 18, 2011 #11

    I like Serena

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    Yep and yep! :)

    So c is the y-intercept of the parabola.

    Huh? :confused:
    I do not understand.
    Where did m = Δy/Δx come from?
    And why would m = (y-c)/x where c is y intercept?

    The tangent line would not have the same y-intercept as the parabola has...

    c is not zero.
    You already concluded it was greater than zero and that is was the y-intercept of the parabola...
  13. Dec 18, 2011 #12
    I already concluded that c=0 because the parabola is not cutting y axis when y=0 and x>0.

    If I am wrong then what shall I do ? Any more hints....:confused:
    Up to now I have found that option A and B are out.

    Hmm so
    where b>0 and c>0
    Now what.. :confused:

    Also b2-4ac >0
    Last edited: Dec 18, 2011
  14. Dec 18, 2011 #13

    I like Serena

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    You can't get more information from y=0 than you already have.
    That is, that there are 2 solutions for y=0, which is expressed in b2-4ac >0.

    Also, you can not conclude that b>0.

    Let's inventorize:
    1. the parabola intersects the positive y-axis at y=c, meaning c>0
    2. the parabola intersects the x-axis at 2 points, meaning b2-4ac >0.
    3. the parabola opens up downward, meaning...?
    4. at which x-coordinate does the parabola have its maximum? And what to deduce?
  15. Dec 18, 2011 #14


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    Earlier you determined that c>0, and you did that correctly. So how can you now take another view, and say c=0?

    Better go back and refresh your mind on the method that led to you getting it right.

  16. Dec 18, 2011 #15
    Hmmm let me think a bit.

    This question was for the standard of class 10th. I can't involve introductory calculus or parabolic equation ( c2 = x2+y2 ) here. I have to do it by applying simple coordinate geometry (up to 10th standard) only.

    Anyways , when parabola is intersecting on positive quadrant of x axis then ,

    x = {-b+-sqrt(b^2-4ac)}/2a , but here I can't say that y=c>0 also. I have got myself totally confused and winded. :confused: If I am taking in account that option B that is c<0 wrong since c>0 and also y>0 then I can also take that in account here. I have got myself totally blundered now !

    The concavity of parabola is downward meaning that y is negative and at one side x > 0 and y<0 AND at other side y<0 and x<0.

    Now what can I do ? :confused:
  17. Dec 18, 2011 #16

    I like Serena

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    Suppose we pick the parabola y=2x2.
    It goes up doesn't it?

    What would it take for the parabola to go down?

    (I like your picture btw. :wink:)
  18. Dec 18, 2011 #17


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    It's fortunate that this image is not in the list of available smileys, or writers here would find it applicable to almost every post on PF!

    A quadrant is an area, one-quarter of the plane. You could phrase this better as, "where parabola intersects the positive x axis then ...".

    You could start by returning to my question which so far has not been answered:
  19. Dec 19, 2011 #18


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    You do overcomplicate the problem. Look at NascentOxygen's big formula:


    and compare with your drawing:
    The curve intersects the y axis at a positive y value.The points on the y axis correspond to x=0. Therefore


    A general parabola can be obtained from the standard y=X2 one by shifting it along x, y and multiplying by a number a as


    If a is a negative number, the parabola is upside down with respect to the standard one. The standard one is open upward. Your parabola is open downward, so

    b/2a shows the shift of the parable along the x direction. As I see, your parabola is not shifted.

  20. Dec 21, 2011 #19
    Concavity of parabola is in the direction of -y axis. In y=2x2 , x2 is positive and so is 2 , so this equation is not satisfied if we compare it with the parabola going downwards as y is negative. You are asking me to use parabolic equation... y=ax2 where x2 is positive and y is negative , so "a" has to be negative.

    This implies a<0 , isn't it ?

    Thanks !
    Answered , I think. See the above reply to ILS' post in my post.

    Ahh !! One more easier method ! Parabola is yielding real and distinct roots. So one value of x is positive and another is of course negative. So product of the two roots is negative as well.

    P or Product of two roots = c/a
    P = c/a
    Since "P" is negative and c>0 so of course a<0 ! :rofl: I couldn't help laughing !

    Can this problem be solved without using parabolic equation ?
    BTW , how can I say that b<0 i.e. option D is wrong ?

    Thanks in advance.
    Last edited: Dec 21, 2011
  21. Dec 21, 2011 #20


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    This is a little exercise for you, to get the feeling about quadratic equations and parabolas. Look at the following quadratic functions and sketch them.
    What is the product of the roots and the sum of the roots of the f(x)=0 quadratic equations? Find a, b, c for all of them. Find the position of the peaks. Discover how the position of the peak is related to the roots.





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