the equation for a firework is h=-5d2+20d+1
h=height above the ground
d=horizontal distance
What would the max height of the firework be?
I am going to assume two things..firstly d2=dē and secondly, that you are in calc
You are attempting to maximize $\displaystyle h(d)=-5d^2+20d+1$
differentiating we get
$\displaystyle h'(d)=-10d+20$
So now we need to find where this equals zero
$\displaystyle h'(d)=0\Rightarrow{-10d+20=0}\Rightarrow{d=2}$
to verify this is a max we use the second derivative test
and since $\displaystyle h''(d)<0,\forall{x}\in\mathbb{R}$
we can see that this is in fact a max
Judging from the questions you posted, this is probably what you want done:
$\displaystyle h=-5d^{2}+20d+1$
Complete the square to get your equation in the form of: $\displaystyle h = a(d - b)^{2} + c$
where the sign of a determines which way the parabola points up (should be negative if you're looking for a max) and (b, c) is your vertex.
The vertex should be the highest point of the parabola and that'll be your answer.
Max value can be determined by finding the x-coordinate of the vertex and then substituting that into the original function:
$\displaystyle y=ax^2+bx+c$
$\displaystyle x=\frac{-b}{2a}$
$\displaystyle f(x)=-5x^2+20x+1$
$\displaystyle x=\frac{-20}{-10}=2$
$\displaystyle f(2)=-5(2)^2+20(2)+1 = 21$ which is the max height.
Or By algebra:
$\displaystyle h=-5d^2+20d+1 = -5(d^2 - 4d) + 1 = -5(d^2 - 4d + 4) + 21 = 21 - 5(d-2)^2$
Now since $\displaystyle (d-2)^2 \geq 0$ always, $\displaystyle h = 21 - 5(d-2)^2 \leq 21$
Equality holds when $\displaystyle d=2$ and the maximum height is $\displaystyle h = 21$