No. of parabolas possible for the given equation.

Question

Given, $f(x)=-x^2+qx+r$. $(q,r) \epsilon R$. $q,r$ are variables. A quadratic equation $f(x)=0$ has a maximum value $m$ ($m$ is a constant) and a root $x=a$. Does $f(x)$ correspond to a unique parabola? Why?

Answer 1

Given, $f(x)=-x^2+qx+r$. $(q,r) \epsilon R$. A quadratic equation $f(x)=0$ has a maximum value $m$

The function might have a maximum. $$ f(x) = -(x^2 - qx - r) = -\left( (x-q/2)^2 -(q^2/4) -r \right) $$ The quadratic term is minimal and with the minus sign maximal, if $x=q/2$. So $f$ has a maximum there. The value is $$ m = (q^2/4) + r \quad (*) $$

and a root $x=a$.

$$ 0 = f(x) = -\left( (x-q/2)^2 -(q^2/4) -r \right) \iff \\ x = \frac{q \pm \sqrt{q^2+4r}}{2} = \frac{q \pm \sqrt{4m}}{2} \quad (**) $$ The number and kind of roots are determined by the term $\Delta = q^2+4r$. If $\Delta = 0$, there is only one root $x=q/2$. If $\Delta > 0$ there are two real roots, if $\Delta <0$ there are no real valued roots (but two complex ones).

Does $f(x)$ correspond to a unique parabola? Why?

If that root $x=a$ is the only root, then the above gives that it is $a=q/2$ and that $q=2a$ from $0 =\Delta = q^2+4r$ we know that $r = -a^2$. Thus both coefficients $q, r$ are uniquely determined as the parabola is too.

If there is a second root or if we do not know the number of roots, and this seems to be the case for this task, then we need more information. As additional information we got the value $m$ of the maximum of $f$, thus $m = f(x^*)$, for some unknown argument $x^*$.

We have two equations, for the maximum we have $(*)$ and for the root we have $(**)$. This leads to:

$$ q = 2a \mp \sqrt{4m} \quad (\#) \\ r = m - (q^2 / 4) $$

If $m = 0$ then we have only one root, and the parabola is uniquely determined.

If $m > 0$, then we have two roots and we do not know which of the two possible signs for the square root in $(\#)$ applies, if the maximum is to the right ($a < x^*$) or to the left of $a$ ($x^* < a$).

So in this case we limit it down to two possible parabolas, with two different coefficients $q, r$, but we can not decide which of the two we are dealing with.

Here are graphs for $a = 2$ and $m = 0$ and $m = 1$:

Answer 2

What is given? What is required?

If $q,r$ are given, then the roots and extreme value are fixed automatically as

$ \dfrac{q \pm \sqrt{q^2+4r}}{2}, (r + q^2/4 ) $ respectively, obtained by standard method of finding roots and extreme values by differentiation.

The two constants $ q,r$ determine only one unique $ y = f(x) = -x^2 + q x + r $ parabola graph.

Do not assign new symbols to dependent quantities $ a, m$ as if they are new or independent!

However,

If $ a, m $ are given

Take the parabola equation in the form $ A y = -(x-a)(x-b) $

Differentiating, $ x_{max}= (a+b)/2 $

Maximum value of parabola is found as

$$ y_m = m = (b-a)^2/(4 A) $$

Find $b$ from above and plug into parabola equation to get one parabola for each value of $A$ chosen.

$$ A y = A f(x) = - (x-a)(x-a-2 \sqrt{m A} ) $$

So the parabolas are not unique like in the first case, they are infinitely many, as seen in the graph where I chose four values for the arbitrary constant $A$.

Treat the problems separately to avoid confusion.

In the example given $ m =1000, a= 1 , (0.05 < A < 0.2) $

EDIT 1:

We can drive home that $A$ is an arbitrary maximum curvature at all top points of same height $m$ for a particular boundary condition chosen. we can formulate a simple differential equation approach as well:

$$ y{''} = -1/A , y ' = -x/A + \tan \alpha , y = - x^2/ (2 A) + x \tan \alpha $$

( arbitrary constant set to $0$ for $x=0,y=0)$.

The $\alpha$ is the variable angle to ground fired by an imaginary gun for the projectile to reach same height $m$. It has two roots

$$ x=0, x = 2 A \tan \alpha ;$$

with maximum height for all $\alpha$

$$ y = y_{max} = m = A \tan {^2}\alpha /2 $$

for all $\alpha$ at $ x= A \tan \alpha.$ The variable $\alpha$ is given by:

$$ \tan \alpha = \sqrt{2 m/A} $$

[We can compare it to a dynamic projectile problem where the vertical component of velocity (proportional to maximum height reached) of a gun shot fired with velocity in various directions].

Answer 3

It need not be a unique parabola. Think geometrically of two possibilities consider a concave down parabola passing through the point $(a,0)$ with maximum value $m$. In one case the left branch passes through $(a,0)$ and in the other case the right branch passes through $(a,0)$. Hence not unique.