Check the picture below.  So the parabola looks more or less like so.
![\bf \textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bhorizontal%20parabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%204p%28x-%20h%29%3D%28y-%20k%29%5E2%20%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5Cqquad%20%5Cstackrel%7Bfocus~point%7D%7B%28h%2Bp%2Ck%29%7D%5Cqquad%20%5Cstackrel%7Bdirectrix%7D%7Bx%3Dh-p%7D%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22p%22~is~negative%7D%7Bop%20ens~%5Csupset%7D%5Cqquad%20%5Cstackrel%7B%22p%22~is~positive%7D%7Bop%20ens~%5Csubset%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \begin{cases} h=-5\\ k=2\\ p=4 \end{cases}\implies 4(4)[x-(-5)]=[y-2]^2\implies 16(x+5)=(y-2)^2 \\\\\\ x+5=\cfrac{1}{16}(y-2)^2\implies x = \cfrac{1}{16}(y-2)^2-5](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20h%3D-5%5C%5C%20k%3D2%5C%5C%20p%3D4%20%5Cend%7Bcases%7D%5Cimplies%204%284%29%5Bx-%28-5%29%5D%3D%5By-2%5D%5E2%5Cimplies%2016%28x%2B5%29%3D%28y-2%29%5E2%20%5C%5C%5C%5C%5C%5C%20x%2B5%3D%5Ccfrac%7B1%7D%7B16%7D%28y-2%29%5E2%5Cimplies%20x%20%3D%20%5Ccfrac%7B1%7D%7B16%7D%28y-2%29%5E2-5)
 
        
                    
             
        
        
        
I'd say maybe by 5:30 b would 
 
        
             
        
        
        
Slope is known as rise over run. Because the line is pointing "\", it is negative.
The rise, the distance from one point to another, specifically from (0,4) to (1,1) is 3, as 4-1=3. your run is 1-0=1.
So your rise over run is -3/1 or, -3.
Your y-int is where when x=0, y=?
In this case y=4 when x=0.
Your equation is
y=-3x+4
 
        
             
        
        
        
Use the pythagoream theorem to check. a^2+b^2=c^2
a=15 b=36 c= 39
225+1296=1521
The equation is true so it is a right triangle.
        
                    
             
        
        
        
The horizontal distance from the firefighter at which the maximum height of water occurs is 10.83m
<u>Explanation:</u>
Given:
h(x) = -0.026x² + 0.563x + 3
where,
h(x) is the height of the water
Initial speed, u = 14m/s
angle, θ = 30°
(a)
Horizontal distance = ?
h(x) = ax² + bx + c
the vertex is found at x = -b/2a
where,
x = distance to the maximum height
As compared to the equation given:
a = -0.026
b = 0.563
Thus, 

Therefore, horizontal distance from the firefighter at which the maximum height of water occurs is 10.83m