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# Math 20-1 question?

A rollercoaster track consists of a series of vertical curves. Near the top of one of these curves, the track enters a tunnel, where passengers ride in the dark for a short duration. The parabola in the graph represents the rollercoaster track. The vertex of this parabola is at (7.5, 21), and another point on the track is at (0, 9).

A steel beam is part of the supporting structure of the rollercoaster track. The beam is attached to the framework at the points (0, 12) and (12, 19).

a. Determine the equations describing the parabola and the line.

b. Using a graphing method, determine the two points where the steel beam is attached to the rollercoaster track. Be sure to include the following:

• any equations you input into the graphing tool

• the window used

• the solutions Relevance

Parabola => y = a(x - 7.5)² + 21

At point (0, 9) we get:

9 = 56.25a + 21

so, a = -16/75

Hence, y = -(16/75)(x - 7.5)² + 21

Line equation is y = 7x/12 + 12

Intercepts occur when:

7x/12 + 12 = -(16/75)(x - 7.5)² + 21

i.e. 175x + 3600 = -64(x² - 15x + 56.25) + 6300

so, 175x + 3600 = -64x² + 960x - 3600 + 6300

=> 64x² - 785x + 900 = 0

Using the quadratic formula we get:

x = (785 ± √385825)/128

so, x = 1.28 or 10.99

i.e. y = 12.75 or 18.41

So, the beam joins the curve at (1.28, 12.75) and at (10.99, 18.41)

The graph below confirms this.

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• You are given two points in a line:  (0, 12) and (12, 19)

Given the general equation for a line:

y = mx + b

Substitute both values for x and y into their own equations to make a system of equations that you can solve:

12 = m(0) + b and 19 = m(12) + b

I'll let you finish solving for "m" and "b" to build your linear equation.

Then for the parabola, vertex form of a quadaratic is:

y = a(x - h)² + k

where the vertex is the point (h, k) which was given to you as (7.5, 21) and we have one other point that can be used for x and y, (0, 9).  Substitute what we know and solve for a:

9 = a(0 - 7.5)² + 21

I'll let you finish that, too.

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