Example 2
Completion requirements
| Example 2 |
An amusement park’s roller coaster has a “hill” with a parabolic shape. The hill begins and ends at the same height, and the horizontal distance between the beginning and end of the hill is \(122 \thinspace \rm{m}\). It reaches a height of \(39 \thinspace \rm{m}\) after a horizontal distance of \(30 \thinspace \rm{m}\). Determine the equation of a quadratic function that represents the height of the hill as a function of the horizontal distance. Set \((0, 0)\) as the beginning of the hill.
Step 1: Draw a diagram and write down what you know.
Step 2: Analyze how to solve the problem.
Because the \(x\)-intercepts are known, the factored form would be a good form to work with. The \(x\)-intercepts are \(0\) and \(122\), so two factors are \((x - 0)\) and \((x - 122)\). However, it is possible that there is also a GCF. As such, the function can be stated as
\(h\left( x \right) = a\left( {x - 0} \right)\left( {x - 122} \right)\)
or
\(h\left( x \right) = ax\left( {x - 122} \right)\), where \(a\) corresponds to the GCF.
Step 3: Determine the value of \(a\).
Substitute the known point \((30, 39)\) into the equation of the function, and solve for \(a\).
\(\begin{align}
h\left( x \right) &= ax\left( {x - 122} \right) \\
39 &= a\left( {30} \right)\left( {30 - 122} \right) \\
39 &= 30a\left( { - 92} \right) \\
39 &= -2760a \\
-\frac{{39}}{{2760}} &= a \\
-\frac{{13}}{{920}} &= a \end{align}\)
Step 4: Make a concluding statement.
The function \(h\left( x \right) = -\frac{{13}}{{920}}x\left( {x - 122} \right)\) represents the height of the hill, \(h(x)\), as a function of the horizontal distance, \(x\).
The function \(h\left( x \right) = -\frac{{13}}{{920}}x\left( {x - 122} \right)\) is not a model for the entire roller coaster, but just for the hill itself. Therefore, the domain must be limited to the horizontal distance that spans the hill.
Domain: {\(x | 0 \le x \le 122, x \in \) R}.
A complete function model of the hill is \(h\left( x \right) = -\frac{{13}}{{920}}x\left( {x - 122} \right)\), \(0 \le x \le 122, x \in\) R.
The
beginning of the hill is at \((0, 0)\). Since the horizontal distance
between the beginning and end of the hill is \(122 \thinspace \rm{m}\), the hill should end
at \((122, 0)\). It was also given that after a horizontal distance of \(30 \thinspace \rm{m}\),
the hill is \(39 \thinspace \rm{m}\) high, so \((30, 39)\) is another point on the graph of the
function.
Step 2: Analyze how to solve the problem.
Because the \(x\)-intercepts are known, the factored form would be a good form to work with. The \(x\)-intercepts are \(0\) and \(122\), so two factors are \((x - 0)\) and \((x - 122)\). However, it is possible that there is also a GCF. As such, the function can be stated as
\(h\left( x \right) = a\left( {x - 0} \right)\left( {x - 122} \right)\)
or
\(h\left( x \right) = ax\left( {x - 122} \right)\), where \(a\) corresponds to the GCF.
Step 3: Determine the value of \(a\).
Substitute the known point \((30, 39)\) into the equation of the function, and solve for \(a\).
\(\begin{align}
h\left( x \right) &= ax\left( {x - 122} \right) \\
39 &= a\left( {30} \right)\left( {30 - 122} \right) \\
39 &= 30a\left( { - 92} \right) \\
39 &= -2760a \\
-\frac{{39}}{{2760}} &= a \\
-\frac{{13}}{{920}} &= a \end{align}\)
Step 4: Make a concluding statement.
The function \(h\left( x \right) = -\frac{{13}}{{920}}x\left( {x - 122} \right)\) represents the height of the hill, \(h(x)\), as a function of the horizontal distance, \(x\).
The function \(h\left( x \right) = -\frac{{13}}{{920}}x\left( {x - 122} \right)\) is not a model for the entire roller coaster, but just for the hill itself. Therefore, the domain must be limited to the horizontal distance that spans the hill.
Domain: {\(x | 0 \le x \le 122, x \in \) R}.
A complete function model of the hill is \(h\left( x \right) = -\frac{{13}}{{920}}x\left( {x - 122} \right)\), \(0 \le x \le 122, x \in\) R.