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Math Help - Quadratic problem

  1. #1
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    Quadratic problem

    Thanks, that was pretty simple

    Find two quadratic functions f and g such that  f (1)  = 0, g(1) = 0 and  f (0) = 10, g(0) = 10 and both have a maximum value of 18.

    So if  f(x) = ax^2 + bx + c then  c= 10 for both f and g.

    I also got  0=a+b+10 for both f and g

    what do I do from here?
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by jgv115 View Post
    Thanks, that was pretty simple

    Find two quadratic functions f and g such that  f (1)  = 0, g(1) = 0 and  f (0) = 10, g(0) = 10 and both have a maximum value of 18.

    So if  f(x) = ax^2 + bx + c then  c= 10 for both f and g.

    I also got  0=a+b+10 for both f and g

    what do I do from here?
    the question basically is that how many quadratic functions f(x)=ax^2+bx+c are there satisfying
    f(1)=0, f(0)=10 and the maximum value attained by f is 18.

    since 1 is a root of f, we can write
    f(x)=k(x-1)(x-p); p being the other root.
    given that f(0)=10, we have kp=10.
    also maximum value of f is 18 and we know that the maxima occurs at x=(p+1)/2 as p and 1 are the roots of f, we have
    f((p+1)/2)=18=(-k/4)((1-p)^2), which simplifies to,
    k((1-p)^2)+72=0. now multiply both sides by k to get,
    (k-kp)^2+72k=0. we already have the value of kp which is 10, so we have,
    (k-10)^2+72k=0 which gives two values of k which are k=-2, k=-50. both are acceptable as k had to be negative( if k is not negative then maxima of f does not exist).
    corresponding values of p are -5 and -0.2.
    so we have two such quadratic functions which can be found out by putting the values of k and p in f(x)=k(x-1)(x-p)
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  3. #3
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by jgv115 View Post
    Thanks, that was pretty simple

    Find two quadratic functions f and g such that  f (1)  = 0, g(1) = 0 and  f (0) = 10, g(0) = 10 and both have a maximum value of 18.

    So if  f(x) = ax^2 + bx + c then  c= 10 for both f and g.

    I also got  0=a+b+10 for both f and g

    what do I do from here?
    to proceed from here you have to exploit on the information that maxima of f is 18.
    you must already be aware that the maxima(or minima) occurs at x=(r1+r2)/2 where r1 and r2 are the roots of the quadratic equation.
    now you must also be knowing that sum of roots can be expressed in terms of the coefficients of powers of x, viz, r1+r2=(-b/a).
    so you will get that maximum of f occurs at x=(-b/2a).
    so use f(-b/2a)=18. from this you will get one more relation in a and b. knowing that a+b+10=0 your problem can be solved.
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