A Parabola's Middle Coefficient

$y = ax^{2} + bx - v$

$\frac{dy}{dx} = 2ax + b$

When $\frac{dy}{dx} = 0$ :

$0 = 2ax + b$

$0 = 2av + b$

$2av = -b$

$v = \frac{-b}{2a}$

When $x=v$ , $y=v$ :

$v = av^{2} + bv - v$

$2v = av^{2} + bv$

$2 = av + b$

$2 = a(\frac{-b}{2a}) + b$

$2 = \frac{-b}{2} + b$

$2 = \frac{b}{2}$

$b = 4$

First consider the y-intercept. At this point $x = 0$ and $y = -v$ where $v \neq 0$ . Substituting $x = 0$ and $y = -v$ in $y = a x^2 + b x + c$ we get $-v = c$ or $v = -c$ . Note that $c \neq 0$ because $v \neq 0$ .

Now again consider the expression $y = a x^2 + b x + c$ where $a \neq 0$ . As this expression represents a parabola we can simplify it to the form $(y - k) = a {x - h}^2$ where $(h, k)$ is the vertex.

By completing the square and rearranging we get

$y - \frac{4ac - b^2}{4a} = a {x - \frac{-b}{2a}}^2$ .

Therefore the vertex is $( \frac{-b}{2a} , \frac{4ac - b^2}{4a} )$

$\frac{-b}{2a} = v = \frac{4ac - b^2}{4a}$

or $\frac{-b}{2a} = -c = \frac{4ac - b^2}{4a}$

taking the left and right equalities separately we obtain

$b = 2ac$ and $b^2 = 8ac$

Substituting the first expression in the second we get $ac = 2$

Therefore $b = 2ac = (2)(2) = 4$ .

Since the vertex of the parabola is at $(v,v)$ , the equation of the parabola can thus be rewritten as $y = a(x-v)^2+v$ .

Substituting the coordinates of the $y$ -intercept $(0,-v)$ into this equation, $-v = a(-v)^2+v$ .

Simplifying, we get $-2v=av^2$ . Thus, $a=\frac{-2}{v}$ .

Expanding the above equation of the parabola, $y = ax^2-2avx+av^2+v$ .

Therefore, equating this with the original equation, $b=-2av=-2\times\frac{-2}{v}\times v=\boxed{4}$ .

Since point(v,v) lies on the parabola, v= av^{ 2 } + bv +c .................. (1) Also, we know that vertex of a parabola is given by x= -b/2a So, -b/2a=v ................... (2) Also, (0,-v) lies on the parabola, which means c = -v . Subs this value in equation (1) to get v= (2-b)/a ................... (3) from equation (2) and (3), eliminate a to get b=4

Clearly, the graph is opening down-wards (since, vertex lies in the first quadrant and $y$ -intercept is negative). Vertex is given $(v,v)$ (lies in the 1st quadrant), hence maximum value of the quadratic polynomial takes place at $x=v$ and the maximum value is $v$ itself.

That is, $av^{2}+bv+c=v$

Also, when $x=0$ , $ax^{2}+bx+c=y$ will be: $c=-v$ ( $y$ -intercept is given)

$i.e.$ $av^{2}+bv-v=v$ $=>av+b=2$ ---- $(*)$

Also, $\frac{-b}{2a}=v$ (given),which gives $\frac{-b}{2v}=a$

Putting value of $a$ in $(*)$ , we get,

$\frac{-bv}{2v}+b=2$ $=>b-\frac{b}{2}=2$ $=>b=\boxed{4}$

Since we have known that the parabola has vertex (v,v),we can simplify it to.Because of y-intercept (0,-v) ,we also got y = ax^{2} + bx - v.It gives us a(x-v)^{2}+v=ax^{2}+bx-v.Expanding this equation,we got ax^{2}-2avx+av^{2}+v=ax^{2}+bx-v.Therefore,we obtain -2av=b and av^{2}+v=-v.So we will get av=-2 and b=4.

Using calculus or method of completing the square,it is found that the vertex has x and y coordinate= $\frac{-b}{2a}$ =v. The y-intercept=c=-v. Given that the coordinate of vertex is (v,v), we knew that at the vertex the equation becomes

v=a $v^{2}$ +bv-v

We get v= $\frac{2-b}{a}$ . Take the value of v from the first sentence above, we get

$\frac{2-b}{a}$ = $\frac{-b}{2a}$

solving for b and get b= $\boxed{4}$

Equation of the parabola with vertex at (v, v) can be written as (y - v) = a(x - v)^2. It also passes through the point (0, -v). This gives the value of a as -2/v. Substitute value of a, expand and compare coeffecients.

We can re-write the equation as (x-v)2=4A(y-v). now plugging in the value of x=0 and y=-v we get the value of A=(-v)/8. then expanding our previous equation and comparing it with the given one gives us the value b=4