Question 1012199
 A parkway 20 meters wide is spanned by a parabolic arc 30 meters long along the horizontal.
 If the parkway is centered, how high must the vertex of the arch be in order to give a minimum clearance of 5 meters over the parkway?
:
that means the 30m arch starts and ends 5m on either side of the 20m highway
From the information given we can obtain the following x,y pairs
x=0; y=0
x=5; y=5
x=25; y=5
Using the form ax^ + bx + c = y; this parabola passes thru the origin, therefore c = 0
Two equations
x=5; y=5
25a + 5b = 5
:
x=25; y=5
625a + 25b = 5
simplify divide by 5
125a + 5b = 1
:
using elimination
125a + 5b = 1
 25a + 5b = 5
---------------------subtraction eliminates b, find a
100a + 0 = -4
a = -4/100
a = -.04
:
Find b using eq 25a + b = 5
-.04(25) + 5b = 5
-1 + 5b = 5
5b = 5 + 1
b = 6/5
b = 1.2
:
the equation: y = -.04x^2 + 1.05x
Looks like this, green line is 5m height
{{{ graph( 300, 200, -5, 32, -5, 18, -.04x^2+1.2x, 5, 9) }}}
:
"how high must the vertex of the arch be" find the axis of symmetry x = -b/(2a)
x = {{{(-1.2)/(2*-.04)}}}
x = 15
Find y when x=15
y = -.04(15^2) + 1.2(15)
y = -9 + 18
Vertex pair: 15, 9
y = 9 meters above the center of the high-way, blue line