Results 1 to 6 of 6

Math Help - Problem finding highest point in quadratic curve

  1. #1
    Member
    Joined
    Jul 2009
    From
    London
    Posts
    109

    Problem finding highest point in quadratic curve

    I have most of this maths puzzle solved but it is the last part which I can't quite work out. Any help would be much appreciated.

    Here is the question:

    The height h metres of a ball at time t seconds after it is thrown up in the air is given by the expression: h = 1 + 15t - 5t^2
    (i) find the times at which the height is 11m.
    (ii) Use your calculator to find the time at which the ball hits the ground.
    (iii) What is the greatest height the ball reaches?

    This is what I have so far:
    (i) -5t^2 + 15t + 1 = 11 (height of 11 m).
    -5t^2 + 15t -10 = 0
    (-5t + 5)(t - 2)
    t = +1 or +2
    ie in 1 second on the way up and 2 seconds on the way down.

    (ii) -5t^2 +15t + 1 = 0 (0 height - ie on ground)
    trickier to factorise so use quadratic equation (-b+/-sqrt(b^2-4as))/2a
    = -15 +/- 7Sqrt(5)
    ----------------
    -10

    t = -0.0652 or 3.0652

    Clearly the negative value is not relevant here so we say the ball hits the ground after 3.0652 seconds.

    (iii) This is the part of the question which I am struggling on. I know that I need to use complete the square technique. Take coefficient - 15, then divide by two then square it - (15/2)^2 = 225/4 or 56.25

    -5t^2 +15t + 1

    Add 224/5 to right side, subtract from left side

    -5t^2 + 15t + 224/4 1-225/4 = -221/4

    I thought the value in the right side - my -221/4, would be the answer - but it is not. What am I doing wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,787
    Thanks
    1570
    To complete the square, the a value needs to be 1.


    -5t^2 + 15t + 1 = -5\left(t^2 - 3t - \frac{1}{5}\right)

     = -5\left[t^2 - 3t + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 - \frac{1}{5}\right]

     = -5\left[\left(t - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{1}{5}\right]

     = -5\left[\left(t - \frac{3}{2}\right)^2 - \frac{45}{20} - \frac{4}{20}\right]

     = -5\left[\left(t - \frac{3}{2}\right)^2 - \frac{49}{20}\right]

     = -5\left(t - \frac{3}{2}\right)^2 + \frac{49}{4}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    If f(x) = ax^2+bx+c, then the maximum/minimum is at \left (-\frac {b}{2a}, -\frac {\Delta}{4a} \right), where \Delta  = b^2-4ac.

    So for f(x) = -5t^2 + 15t + 1, we have \left (-\frac {b}{2a}, -\frac {\Delta}{4a} \right) = \left (\frac {15}{10}, \frac {15^2+20}{20} \right) = \left(\frac{3}{2}, \frac {245}{20} \right) = \left (\frac {3}{2}, \frac {49}{4} \right).
    Last edited by TheCoffeeMachine; September 24th 2010 at 02:46 AM. Reason: Fixed the 'minus' (as pointed out below).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,787
    Thanks
    1570
    Quote Originally Posted by TheCoffeeMachine View Post
    If f(x) = ax^2+bx+c, then the maximum/minimum is at \left (-\frac {b}{2a}, -\frac {\Delta}{4a} \right), where \Delta  = b^2-4ac.

    So for f(x) = -5t^2 + 15t + 1, we have \left (-\frac {b}{2a}, -\frac {\Delta}{4a} \right) = \left (-\frac {15}{10}, \frac {15^2+20}{20} \right) = \frac {245}{20} \right) = \left (-\frac {3}{2}, \frac {49}{4} \right).
    I believe you'll find it's actually (x,y) = \left(\frac{3}{2}, \frac{49}{4}\right)...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    433
    Thanks
    12
    Quote Originally Posted by Prove It View Post
    I believe you'll find it's actually (x,y) = \left(\frac{3}{2}, \frac{49}{4}\right)...
    His working was correct, he just didn't get the final answer correct.

    It should have been:

    \left (-\frac {b}{2a}, -\frac {\Delta}{4a} \right) = \left (-\frac {15}{-10}, \frac {15^2+20}{20} \right) = \frac {245}{20} \right) = \left (\frac {3}{2}, \frac {49}{4} \right).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jul 2009
    From
    London
    Posts
    109
    Why does a have to be 1?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 30th 2011, 02:13 PM
  2. Replies: 2
    Last Post: June 10th 2011, 02:20 PM
  3. Minimum point on a quadratic curve
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 4th 2010, 11:47 AM
  4. Replies: 3
    Last Post: October 1st 2009, 05:35 AM
  5. Replies: 4
    Last Post: December 23rd 2008, 06:59 AM

Search Tags


/mathhelpforum @mathhelpforum