assume an arch is shaped like a parabola. It is 20ft wide at base and 80 ft high
a. find equation that models arch
b. how wide is arch 20ft above ground?
+36916 Let's put one base leg at 0,0 and the other leg at 20,0
and write the equation of a parabola in vertex form
the vertex will be at (h,k) = 10,80
Vertex from = a(x-h)^2 + k so:
f(x) = a (x-10)^2 + k
= a (x^2-20x +100) + 80
We know that point 0,0 is on the graph, so let's sub that in to help find 'a'
0 = a(0^2-20(0) +100) + 80
0 = 100a +80
a = -80/100 = -4/5
So the equation is f(x) = -4/5 (x-10)^2 + 80
at f(x) = 20 20 = -4/5(x-10)^2 + 80
60 = 4/5 (x-10)^2
75 = (x-10)^2
+- sqrt 75 = x-10
x = sqrt 75 +10 and - sqrt 75 + 10
x = 18.66 and 1.34 ft The width is the difference between these x coordinates = 17.32 ft
+128475 Thanks EP....here's another way using symmetry about the y axis
a. Let the ends of the arch be at (-10, 0) and (10,0)
Let the apex of the arch be at (0,80)
So....we just need to find "a" in this function
y = ax^2 + 80
We know the point (10, 0) = (x, y) is on the graph...so....filling in what we know, we have
0 = a(10)^2 + 80
0 = 100a + 80 subtract 80 from both sides
-80 = 100a divide both sides by 100
-80/ 100 = a = - 4/5
So....the function is
y = (-4/5)x^2 + 80
b. At 20 ft above the ground, we have
20 = (-4/5)x^2 + 80 subtract 80 from both sides
-60 = (-4/5)x^2 multiply both sides by -5/4
75 = x^2 take the positve root
sqrt (75) = x = 5sqrt(3) ft
But this is only half the width...so multiplying by 2 we have
Width = 10sqrt(3) ≈ 17.3 ft
