Results 1 to 3 of 3

Math Help - find values for x

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    72

    Post find values for x

    Hi, I have a problem that trying to solve all evning and it is in two parts (a) If a is a positive constant, find the set of valus of x for which a(x^2 +2x-8) is negative. Find the value of a if this function has a minimum value of -27.
    (b) Find two quadratic funcions of x which are zero at x=1, which take the value of 10 when x=0 and which have a maximum value of 18. Sketch the graps of these two functions. Can ou help me with this problem?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,888
    Thanks
    326
    Awards
    1
    Quote Originally Posted by scrible View Post
    If a is a positive constant, find the set of valus of x for which a(x^2 +2x-8) is negative. Find the value of a if this function has a minimum value of -27.
    If a is positive then we need to find where x^2 + 2x - 8 < 0. Start by finding out where the trinomial is zero. I get these to be at x = -4 and x = 2. It is simple to see that the negative range for this is -4 < x < 2.

    Can you get the rest from here? (Hint: Consider where the parabola has its absolute minimum.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,888
    Thanks
    326
    Awards
    1
    Quote Originally Posted by scrible View Post
    (b) Find two quadratic funcions of x which are zero at x=1, which take the value of 10 when x=0 and which have a maximum value of 18. Sketch the graps of these two functions.
    You are speaking of a quadratic. The general form is
    y = ax^2 + bx + c

    For x = 1 we have a zero, so y = 0.
    0 = a + b + c

    When x = 0 we have y = 10, so
    10 = c

    So far the quadratic is
    y = ax^2 + bx + 10
    where a + b = -10

    Now, we know that the quadratic has a maximum of 18. This means that a is negative. The last step is a bit involved, but if you recall,
    x_{max/min} = -\frac{b}{2a}
    You can plug that into the y = ax^2 + (-a - 10)x + 10 equation to get
    18 = ax^2 + (-a - 10)x + 10
    (Recall that the maximum y is 18.) It's tedious but you will get a quadratic in a that give you your two solutions.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find all possible values
    Posted in the Algebra Forum
    Replies: 5
    Last Post: September 19th 2013, 04:52 AM
  2. Replies: 2
    Last Post: January 28th 2010, 01:39 AM
  3. help with find values
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 21st 2009, 02:54 AM
  4. Find the values of F1 and F2
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: August 19th 2009, 10:20 AM
  5. Replies: 1
    Last Post: May 24th 2009, 05:16 AM

Search Tags


/mathhelpforum @mathhelpforum