Precalculus word problem

Question

An arch going into an alley is in the shape of a downward opening parabola. The base of the arch is 10 ft. wide, and the arch is 15 ft. high at its center. How wide is the arch 10 ft. up?

Answer 1

Since the arch is a downward facing Parabola and you're given the width of the base (10ft) as well as the height at the center (15ft) you can use the Vertex Form equation for a parabola y=a(x-h)^2+k.

To help, you can draw an xy-axis then label your x-intercepts at the points (0,0) and (10,0), then your vertex is at the point (5,15).

To find the correct equation you'll need to solve for a. So, plug in the coordinates of your vertex (5,15) as h=5 and k=15. Now pick either one of your intercept points, for example I used the point (0,0) and substitute the values into your Vertex Form equation like so 0=a(0-5)^2+15. Now, simplify and solve for a and you should get a=-3/5.

Next, to find the width at 10ft up you'll need to first set the equation of your parabola equal to 10 then solve for x₁ and x₂.

10 = -3/5 * (x-5)^2 +15 which yields the values x₁ =2.113 and x₂=7.887.

Finally, the width of the arch relative to those x-values is found by simply adding the distance from the center (x=5) to each x₁ =2.113 and x₂=7.887. Do this once, then use symmetry

5 - 2.113 = 2.887 therefore the distance between x₁ and x₂ is roughly 5.774 ft.

Therefore the width of the arch 10 feet up is about 5.8 ft.

Answer 2

Draw and label a diagram!

Place vertex at (0, 15), x-intercepts at (-5, 0) and (5, 0)

f(x) = a(x - 0)² + 15

f(x) = ax² + 15

0 = a(5)² + 15

-15 = 25a

-3/5 = a

f(x) = -3/5(x - 0)² + 15

10 = -3x²/5 + 15

-5 = -3x²/5

-25 = -3x²

25/3 = x²

±(5√3)/3 = x

Distance between arches at 10 feet up is 10√3 / 3 feet.