Real-World Applications: Completing the Square for Maximum Height and Range
Grade 9 students in California Reveal Math Algebra 1 learn to apply completing the square to real-world projectile and fountain problems to find maximum height and landing distance. For a model h(t)=at²+bt+c with a<0, complete the square to reach vertex form h(t)=a(t-h)²+k — the maximum height is k at time t=h. To find when the object lands, set vertex form equal to zero, apply the Square Root Property, and keep only the positive time solution. For example, h(t)=-16t²+32t+6 becomes -16(t-1)²+22, giving maximum 22 feet at t=1 second, landing at t≈2.17 seconds.
Key Concepts
For a real world quadratic model $h(t) = at^2 + bt + c$ (where $h$ is height and $t$ is time), convert to vertex form by completing the square:.
$$h(t) = a(t h)^2 + k$$.
Common Questions
How do you find maximum height using completing the square?
Rewrite the quadratic model in vertex form by completing the square. The maximum height equals k (the y-coordinate of the vertex), occurring at time t=h.
How do you find when a projectile lands?
Set vertex form equal to zero and solve using the Square Root Property. Keep only the positive time solution since negative time is not physically meaningful.
Can you show the full solution for h(t)=-16t²+32t+6?
Factor out -16: -16(t²-2t)+6. Complete the square: -16(t-1)²+22. Maximum height is 22 feet at t=1 second. Setting equal to zero: (t-1)²=22/16=11/8, so t=1+sqrt(11/8)≈2.17 seconds.
How do you solve a fountain problem using completing the square?
For h(x)=-0.5x²+3x: factor to -0.5(x²-6x)=-0.5(x-3)²+4.5. Maximum height is 4.5 feet at x=3 feet, and the water lands at x=6 feet when h(x)=0.
Why must a be negative for a maximum to exist?
The maximum only exists when the parabola opens downward, which requires a<0. If a>0 the parabola opens upward and has a minimum, not a maximum.
Which unit covers this completing the square application?
This skill is from Unit 10: Quadratic Functions in California Reveal Math Algebra 1, Grade 9.