Solving Systems of Equations Algebraically?

A canopy tour company designs zip line rides through forest and jungle environments. One particular zip line takes the passenger through a waterfall, the maximum height of which is 167 ft. The dimensions of both the zip line and the waterfall are shown in the image below. 

Model this scenario with a system of equations.

Hint, use:

y = mx + b

and

y = a(x -p)2 + q

Attachment image

1 Answer

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  • 3 months ago

    Figure out an equation for both functions.

    Let's set the origin (0,0) in the lower left.

    We have a line with a negative slope going down 130 feet in a horizontal distance of 100 feet.

    m = -130/100

    m = -1.3

    And the y-intercept is 130 feet.

    y = -1.3x + 130

    For the parabola, it appears the vertex is at 167 feet when x = 0, so the vertex is at (0,167)

    Using the vertex form y = a(x - p)² + q, plug in the vertex point (p,q) = (0,167)

    y = ax² + 167

    Next we have to figure out a, so that y = 0 when x = 75:

    0 = a(75²) + 167

    0 = 5625a + 167

    -5625a = 167

    a = 167/-5625

    y = -(167/5625)x² + 167

    You didn't say if you were supposed to figure out the intersection, but you now have the two equations to be able to do that.

    One way would be to equate them:-(167/5625)x² + 167 = -1.3x + 130

    Then get everything on one side:

    -(167/5625)x² + 1.3x + 37 = 0

    Then use the quadratic formula.

    You want the positive x value:

    x ≈ 63.434

    And plug that into either equation to find y:

    y = -1.3(63.434) + 130

    y ≈ 47.536

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