Question 1198183: A train tunnel is modeled by the quadratic function h(x)= - 0.2x^2+ 27 where x is the distance in feet from the center of the tracks and h (x) is the height of the tunnel. Also in feet. Assume that the high point of the tunnel is directly in line with the center of the train tracks. What is the maximum height of the tunnel. Feet how wide is the base of the tunnel. Feet
Answer by math_tutor2020(3817)   (Show Source):
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Rewrite y = -0.2x^2+ 27 as y = -0.2(x-0)^2+ 27
The second equation is of the form
y = a(x-h)^2 + k
where,
a = -0.2
h = 0
k = 27
The vertex is (h,k) = (0,27) representing the highest point on this parabola.
We know the parabola opens downward because a < 0.
Therefore, the highest point on the tunnel is at 27 feet.
Put another way: the largest h(x) can get is h(x) = 27, which occurs when x = 0.
We'll plug in y = 0 to solve for x, so we can find the roots of this parabola. Doing this will then help us determine how wide the tunnel is at the base.
y = -0.2x^2+ 27
0 = -0.2x^2+ 27
0.2x^2 = 27
x^2 = 27/0.2
x^2 = 135
x = sqrt(135) or x = -sqrt(135)
x = 11.61895 or x = -11.61895
Those decimal values are approximate.
Use a graphing tool like Desmos to confirm the answers.
The distance between the roots is
|11.61895-(-11.61895)| = |11.61895+11.61895| = 23.2379 feet approximately.
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Answers:
Max height of tunnel = 27 feet
width at the base = about 23.2379 feet
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