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Completion requirements
To score a \(3\rm{-point}\) field goal in the game of football, the
place kicker must kick the ball between the uprights and above the \(10 \thinspace \rm{ft}\) high cross-bar, which are located in the end zone of the
football field.
A ball is positioned \(43 \thinspace \rm{yd}\) from the end zone, where the place kicker will attempt to kick a field goal.
The flight of the ball is parabolic, the maximum height reached by the ball is \(23 \thinspace \rm{yd}\) above the turf, and the ball lands \(1.5 \thinspace \rm{yd}\) behind the uprights.
Will the football clear the cross-bar and be a successful field goal?
A ball is positioned \(43 \thinspace \rm{yd}\) from the end zone, where the place kicker will attempt to kick a field goal.
The flight of the ball is parabolic, the maximum height reached by the ball is \(23 \thinspace \rm{yd}\) above the turf, and the ball lands \(1.5 \thinspace \rm{yd}\) behind the uprights.
Will the football clear the cross-bar and be a successful field goal?
Step 1: Write down what you know.
Mark the point where the football starts as \((0, 0)\) and the point where the football lands as \((44.5, 0)\).
The vertex of the quadratic function is \((22.25, 23)\) because the maximum occurs at the half-way mark.
Step 2: Analyze how to solve the problem.
Because both \(x\)-intercepts and the vertex are known, both forms of a quadratic function can be used. The following solution will begin with factored form.
\(x\)-intercepts: \((0, 0)\) and \((44.5, 0)\)
\(\begin{array}{l}
f\left( x \right) = a\left( {x - 0} \right)\left( {x - 44.5} \right) \\
f\left( x \right) = ax\left( {x - 44.5} \right) \\
\end{array}\)
Step 3: Determine the value of \(a\).
Substitute a known point (the vertex) into the equation of the function, and solve for \(a\).
\(\begin{align}
f\left( x \right) &= a\left( {x - 0} \right)\left( {x - 44.5} \right) \\
23 &= a\left( {22.25} \right)\left( {22.25 - 44.5} \right) \\
23 &= -495.0625a \\
-0.046... &= a \end{align}\)
Step 4: Write the equation of the quadratic function in standard form.
\(\begin{array}{l}
f\left( x \right) = \left( { -0.046...} \right)x\left( {x - 44.5} \right) \\
f\left( x \right) = -0.046...x^2 + 2.067...x \end{array}\)
Step 5: Use this function to solve the problem.
The height of the cross-bar between the uprights is \(10 \thinspace \rm{ft}\). So, \(43 \thinspace \rm{yd}\) from the point labelled \((0, 0)\), the football must be at least \(10 \thinspace \rm{ft}\) in the air in order to be deemed a field goal.
Note: in order to work with this function \(f\left( x \right) = -0.046...x^2 + 2.067...x\), all measurements must be in yards. As such, convert the height of the \(10 \thinspace \rm{ft}\) high cross-bar to yards.
Step 6: Conclusion
In conclusion, the football will not clear the uprights, and thus a field goal will not be scored. The vertical height of the ball at \(x = 43 \thinspace \rm{yd}\) is \(2.996…\thinspace \rm{yd}\), which is not over the required \(3.333… \thinspace \rm{yd}\).
Mark the point where the football starts as \((0, 0)\) and the point where the football lands as \((44.5, 0)\).
The vertex of the quadratic function is \((22.25, 23)\) because the maximum occurs at the half-way mark.
Step 2: Analyze how to solve the problem.
Because both \(x\)-intercepts and the vertex are known, both forms of a quadratic function can be used. The following solution will begin with factored form.
\(x\)-intercepts: \((0, 0)\) and \((44.5, 0)\)
\(\begin{array}{l}
f\left( x \right) = a\left( {x - 0} \right)\left( {x - 44.5} \right) \\
f\left( x \right) = ax\left( {x - 44.5} \right) \\
\end{array}\)
Step 3: Determine the value of \(a\).
Substitute a known point (the vertex) into the equation of the function, and solve for \(a\).
\(\begin{align}
f\left( x \right) &= a\left( {x - 0} \right)\left( {x - 44.5} \right) \\
23 &= a\left( {22.25} \right)\left( {22.25 - 44.5} \right) \\
23 &= -495.0625a \\
-0.046... &= a \end{align}\)
Step 4: Write the equation of the quadratic function in standard form.
\(\begin{array}{l}
f\left( x \right) = \left( { -0.046...} \right)x\left( {x - 44.5} \right) \\
f\left( x \right) = -0.046...x^2 + 2.067...x \end{array}\)
Step 5: Use this function to solve the problem.
The height of the cross-bar between the uprights is \(10 \thinspace \rm{ft}\). So, \(43 \thinspace \rm{yd}\) from the point labelled \((0, 0)\), the football must be at least \(10 \thinspace \rm{ft}\) in the air in order to be deemed a field goal.
Note: in order to work with this function \(f\left( x \right) = -0.046...x^2 + 2.067...x\), all measurements must be in yards. As such, convert the height of the \(10 \thinspace \rm{ft}\) high cross-bar to yards.
\(\begin{align}
\frac{{3{\rm{ ft}}}}{{1{\rm{ yd}}}} &= \frac{{10{\rm{ ft}}}}{x} \\
3{\rm{ ft }}{\rm{ }}x &= (10{\rm{ ft }})({\rm{ 1 yd}}) \\
x &= \frac{{(10{\rm{ ft )}}{\rm{ (1 yd)}}}}{{3{\rm{ ft}}}} \\
x &= 3.333...{\rm{ yd}} \\
\end{align}\)
\frac{{3{\rm{ ft}}}}{{1{\rm{ yd}}}} &= \frac{{10{\rm{ ft}}}}{x} \\
3{\rm{ ft }}{\rm{ }}x &= (10{\rm{ ft }})({\rm{ 1 yd}}) \\
x &= \frac{{(10{\rm{ ft )}}{\rm{ (1 yd)}}}}{{3{\rm{ ft}}}} \\
x &= 3.333...{\rm{ yd}} \\
\end{align}\)
Determine the value of the function (height of the ball) when \(x = 43 \thinspace \rm{yd}\).
\(\begin{array}{l}
f\left( x \right) = -0.046...x^2 + 2.067...x \\
f\left( {43} \right) = -0.046...\left( {43} \right)^2 + 2.067...\left( {43} \right) \\
f\left( {43} \right) = 2.996... \end{array}\)
\(\begin{array}{l}
f\left( x \right) = -0.046...x^2 + 2.067...x \\
f\left( {43} \right) = -0.046...\left( {43} \right)^2 + 2.067...\left( {43} \right) \\
f\left( {43} \right) = 2.996... \end{array}\)
Step 6: Conclusion
In conclusion, the football will not clear the uprights, and thus a field goal will not be scored. The vertical height of the ball at \(x = 43 \thinspace \rm{yd}\) is \(2.996…\thinspace \rm{yd}\), which is not over the required \(3.333… \thinspace \rm{yd}\).