1. ## the firework

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?

2. Originally Posted by mcgill33
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 $h(d)=-5d^2+20d+1$

differentiating we get

$h'(d)=-10d+20$

So now we need to find where this equals zero

$h'(d)=0\Rightarrow{-10d+20=0}\Rightarrow{d=2}$

to verify this is a max we use the second derivative test

and since $h''(d)<0,\forall{x}\in\mathbb{R}$
we can see that this is in fact a max

3. Judging from the questions you posted, this is probably what you want done:

$h=-5d^{2}+20d+1$

Complete the square to get your equation in the form of: $h = a(d - b)^{2} + c$

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.

4. Max value can be determined by finding the x-coordinate of the vertex and then substituting that into the original function:

$y=ax^2+bx+c$

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

$f(x)=-5x^2+20x+1$

$x=\frac{-20}{-10}=2$

$f(2)=-5(2)^2+20(2)+1 = 21$ which is the max height.

5. Originally Posted by mcgill33
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?
Or By algebra:

$h=-5d^2+20d+1 = -5(d^2 - 4d) + 1 = -5(d^2 - 4d + 4) + 21 = 21 - 5(d-2)^2$

Now since $(d-2)^2 \geq 0$ always, $h = 21 - 5(d-2)^2 \leq 21$

Equality holds when $d=2$ and the maximum height is $h = 21$