Quadratic problem

**jgv115** · January 15th 2011, 08:29 PM

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?

**abhishekkgp** · January 15th 2011, 09:10 PM

Originally Posted by jgv115

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 $f(x)=k(x-1)(x-p)$

**abhishekkgp** · January 15th 2011, 09:21 PM

Originally Posted by jgv115

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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