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    Quadratic problem

    Thanks, that was pretty simple

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

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

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

    what do I do from here?
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    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$\displaystyle f (1) = 0, g(1) = 0 $ and $\displaystyle f (0) = 10, g(0) = 10 $ and both have a maximum value of 18.

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

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

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

    since 1 is a root of $\displaystyle f$, we can write
    $\displaystyle f(x)=k(x-1)(x-p)$; p being the other root.
    given that $\displaystyle f(0)=10$, we have $\displaystyle kp=10$.
    also maximum value of $\displaystyle f$ is 18 and we know that the maxima occurs at $\displaystyle x=(p+1)/2$ as p and 1 are the roots of f, we have
    $\displaystyle f((p+1)/2)=18=(-k/4)((1-p)^2)$, which simplifies to,
    $\displaystyle k((1-p)^2)+72=0$. now multiply both sides by k to get,
    $\displaystyle (k-kp)^2+72k=0$. we already have the value of $\displaystyle kp$ which is 10, so we have,
    $\displaystyle (k-10)^2+72k=0$ which gives two values of k which are $\displaystyle k=-2, k=-50$. 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 $\displaystyle f(x)=k(x-1)(x-p)$
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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$\displaystyle f (1) = 0, g(1) = 0 $ and $\displaystyle f (0) = 10, g(0) = 10 $ and both have a maximum value of 18.

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

    I also got $\displaystyle 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 $\displaystyle f$ is 18.
    you must already be aware that the maxima(or minima) occurs at $\displaystyle x=(r1+r2)/2$ where $\displaystyle r1$ and $\displaystyle r2$ 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, $\displaystyle r1+r2=(-b/a)$.
    so you will get that maximum of f occurs at $\displaystyle x=(-b/2a)$.
    so use $\displaystyle f(-b/2a)=18$. from this you will get one more relation in a and b. knowing that $\displaystyle a+b+10=0$ your problem can be solved.
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