\(y = -2(x + 3)^2 + 8\)
Vertex Form: \(y = -2(x + 3)^2 + 8\)
Vertex \((h, k)\): \((-3, 8)\)
Axis of Symmetry: \(x = -3\)
y-intercept: \((0, -10)\)
6. \(y = \frac{1}{2}x^2 - 4x + 6\) \(a = \frac{1}{2}, \; b = -4, \; c = 6\)
Algebraic Derivation:
\(y = \frac{1}{2}(x^2 - 8x) + 6\) [factor out \(\frac{1}{2}\): \(-4 \div \frac{1}{2} = -8\)]
\(y = \frac{1}{2}(x^2 - 8x + \mathbf{16}) + 6 - \mathbf{(\frac{1}{2})(16)} = \frac{1}{2}(x - 4)^2 + 6 - 8\)
\(y = \frac{1}{2}(x - 4)^2 - 2\)
Vertex Form: \(y = \frac{1}{2}(x - 4)^2 - 2\)
Vertex \((h, k)\): \((4, -2)\)
Axis of Symmetry: \(x = 4\)
y-intercept: \((0, 6)\)
Review for Quiz • Vertex Form and Quadratic Applications • Teacher Answer Key Page 2 of 12
Full algebraic completion of the square, complete coordinates table, and high-resolution plotted parabola.
Opens Upward
1. Completed Square Derivation:
\(y = (x^2 - 4x + \mathbf{4}) + 1 - \mathbf{4}\)
\(y = (x - 2)^2 - 3\)
Vertex Form: \(y = (x - 2)^2 - 3\)
Vertex \((h, k)\): \((2, -3)\)
Axis of Symmetry: \(x = 2\)
y-intercept: \((0, 1)\)
2. Strategic Coordinates Around Vertex:
\(x\)01234
\(y\) 1 -2 -3 -2 1
Domain:
\((-\infty, \infty)\)
Range:
\([-3, \infty)\)
Plotted Solution Curve:
x y Vertex (2, -3)
✓ Minimum Value = -3 • Axis of Symmetry: \(x = 2\)
Review for Quiz • Vertex Form and Quadratic Applications • Teacher Answer Key Page 3 of 12
Parabola with negative leading coefficient \(a = -2\), stretched vertically, opening downward.
Opens Downward
1. Completed Square Derivation:
\(y = -2(x^2 - 4x + \mathbf{4}) - 5 - \mathbf{(-2)(4)}\)
\(y = -2(x - 2)^2 - 5 + 8\)
\(y = -2(x - 2)^2 + 3\)
Vertex Form: \(y = -2(x - 2)^2 + 3\)
Vertex \((h, k)\): \((2, 3)\)
Axis of Symmetry: \(x = 2\)
Maximum Value: \(y = 3\)
2. Strategic Coordinates Around Vertex:
\(x\)01234
\(y\) -5 1 3 1 -5
Domain:
\((-\infty, \infty)\)
Range:
\((-\infty, 3]\)
Plotted Solution Curve:
x y (2, 3)
✓ Maximum Value = 3 • Axis of Symmetry: \(x = 2\)
Review for Quiz • Vertex Form and Quadratic Applications • Teacher Answer Key Page 4 of 12
Solid boundary curve with solution region shaded inside and above the parabola.
Solid Boundary
1. Vertex Form Conversion:
\(y \ge (x^2 + 2x + \mathbf{1}) - 3 - \mathbf{1}\)
\(y \ge (x + 1)^2 - 4\)
Vertex: \((-1, -4)\)
Boundary Line: Solid Line (\(\ge\))
Axis of Symmetry: \(x = -1\)
y-intercept: \((0, -3)\)
2. Test Point Verification: \((0, 0)\)
\(0 \ge 0^2 + 2(0) - 3 \implies 0 \ge -3\) ✓ TRUE
Because \((0,0)\) yields a true statement, the solution region is INSIDE / ABOVE the parabola curve containing the test point.
Plotted Solution & Shaded Region:
x y (0,0) ✓ (-1, -4)
✓ Solid Boundary Curve • Shaded Interior Region
Review for Quiz • Vertex Form and Quadratic Applications • Teacher Answer Key Page 5 of 12
Dashed boundary curve with solution region shaded below and inside the downward parabola.
Dashed Boundary
1. Vertex Form Conversion:
\(y < -(x^2 - 6x + \mathbf{9}) - 5 - \mathbf{(-1)(9)}\)
\(y < -(x - 3)^2 - 5 + 9\)
\(y < -(x - 3)^2 + 4\)
Vertex: \((3, 4)\)
Boundary Line: Dashed Line (\(<\))
Axis of Symmetry: \(x = 3\)
y-intercept: \((0, -5)\)
2. Test Point Verification: \((3, 0)\)
\(0 < -(3 - 3)^2 + 4 \implies 0 < 4\) ✓ TRUE
Because \((3,0)\) yields a true statement, the solution region is INSIDE / BELOW the parabola containing the vertex axis interior.
Plotted Solution & Shaded Region:
x y (3,0) ✓ (3, 4)
✓ Dashed Boundary Curve • Shaded Interior Region
Review for Quiz • Vertex Form and Quadratic Applications • Teacher Answer Key Page 6 of 12
Formulating free-fall equations from initial height (\(v_0 = 0\)).
Problems 11 & 12
11. [Dropped Item] Cell Tower Tool Drop Model: \(h(t) = -16t^2 + 144\)
Student Model Setup: Since dropped with no initial push, \(v_0 = 0\), \(h_0 = 144\text{ ft}\) \(\implies h(t) = -16t^2 + 144\).
a) Height at \(t = 2\) seconds:
\(h(2) = -16(2)^2 + 144\)
\(h(2) = -16(4) + 144 = -64 + 144\)
\(h(2) = 80\text{ feet}\)
b) Time to strike ground (\(h(t) = 0\)):
\(0 = -16t^2 + 144 \implies 16t^2 = 144\)
\(t^2 = 9 \implies t = \pm 3\)
\(t = 3\text{ seconds}\) (\(t > 0\))
12. [Dropped Item] Cargo Supply Crate Model: \(h(t) = -16t^2 + 400\)
Student Model Setup: Hovering helicopter implies \(v_0 = 0\), initial height \(h_0 = 400\text{ ft}\) \(\implies h(t) = -16t^2 + 400\).
a) Time to reach the ground (\(h(t) = 0\)):
\(0 = -16t^2 + 400 \implies 16t^2 = 400\)
\(t^2 = \frac{400}{16} = 25 \implies t = 5\)
\(t = 5\text{ seconds}\)
b) Time when height is 256 feet:
\(256 = -16t^2 + 400 \implies 16t^2 = 144\)
\(t^2 = 9 \implies t = 3\)
\(t = 3\text{ seconds}\)
Review for Quiz • Vertex Form and Quadratic Applications • Teacher Answer Key Page 7 of 12
Student formulation of upward initial velocity models.
Problems 13 & 14
13. [Projected Upward] Goalkeeper Punt Model: \(h(t) = -16t^2 + 64t\)
Student Model Setup: Ground level \(h_0 = 0\), initial upward velocity \(v_0 = 64\text{ ft/s}\) \(\implies h(t) = -16t^2 + 64t\).
a) Vertex Form:
\(h(t) = -16(t^2 - 4t + 4) + 64\)
\(h(t) = -16(t - 2)^2 + 64\)
b) Max Height & Time:
Vertex is \((2, 64)\)
\(64\text{ ft}\) at \(t = 2\text{ s}\)
c) Landing Time (\(h = 0\)):
\(-16t(t - 4) = 0\)
\(t = 4\text{ seconds}\)
14. [Projected Upward] Rocket Platform Launch Model: \(h(t) = -16t^2 + 96t + 10\)
Student Model Setup: Elevated platform \(h_0 = 10\text{ ft}\), initial upward velocity \(v_0 = 96\text{ ft/s}\) \(\implies h(t) = -16t^2 + 96t + 10\).
a) Time to Peak and Maximum Height:
\(t = -\frac{b}{2a} = -\frac{96}{2(-16)} = 3\text{ seconds}\)
\(h(3) = -16(9) + 96(3) + 10 = -144 + 288 + 10\)
Max Height = \(154\text{ feet at } t = 3\text{ s}\)
b) Touchdown Time (\(h = 0\)):
\(t = \frac{-96 \pm \sqrt{96^2 - 4(-16)(10)}}{-32} = \frac{-96 \pm \sqrt{9856}}{-32}\)
\(t = \frac{-96 - 99.277}{-32} = \frac{-195.277}{-32}\)
\(t \approx 6.10\text{ seconds}\)
Review for Quiz • Vertex Form and Quadratic Applications • Teacher Answer Key Page 8 of 12
Student formulation of model and discriminant evaluation.
Problems 15 & 16
15. [Projected Upward] Archery Elevation Model: \(h(t) = -16t^2 + 112t + 6\)
Student Model Setup: Stand height \(h_0 = 6\text{ ft}\), initial velocity \(v_0 = 112\text{ ft/s}\) \(\implies h(t) = -16t^2 + 112t + 6\).
a) Equation for \(h(t) = 170\):
\(-16t^2 + 112t + 6 = 170\)
\(-16t^2 + 112t - 164 = 0\)
\(4t^2 - 28t + 41 = 0\)
b) Solve and Physical Meaning:
\(t = \frac{28 \pm \sqrt{784 - 4(4)(41)}}{8} = \frac{28 \pm \sqrt{128}}{8}\)
\(t_1 \approx 2.09\text{ s}\) (ascending) • \(t_2 \approx 4.91\text{ s}\) (descending)
Two positive times exist because the arrow crosses 170 ft on the way up and again on the way down.
16. [Discriminant] Baseball Catwalk Model: \(h(t) = -16t^2 + 80t + 3\)
Student Model Setup: Bat height \(h_0 = 3\text{ ft}\), initial velocity \(v_0 = 80\text{ ft/s}\) \(\implies h(t) = -16t^2 + 80t + 3\).
a) Standard form (\(h = 110\)):
\(-16t^2 + 80t + 3 = 110\)
\(-16t^2 + 80t - 107 = 0\)
b) Compute \(\Delta = b^2 - 4ac\):
\(\Delta = 80^2 - 4(-16)(-107)\)
\(\Delta = 6400 - 6848\)
\(\Delta = -448 < 0\)
c) Catwalk Conclusion:
Does NOT hit catwalk.
Since \(\Delta < 0\), there are no real solutions. Max height is only \(103\text{ ft}\), below \(110\text{ ft}\).
Review for Quiz • Vertex Form and Quadratic Applications • Teacher Answer Key Page 9 of 12
Student model formulations and consecutive integer parameter solving.
Problems 17 & 18
17. [Discriminant] Firework Safety Altitude Model: \(h(t) = -16t^2 + 68t + 24\)
Student Model Setup: Roof height \(h_0 = 24\text{ ft}\), initial velocity \(v_0 = 68\text{ ft/s}\) \(\implies h(t) = -16t^2 + 68t + 24\).
a) Standard Form (\(h = 100\)):
\(-16t^2 + 68t + 24 = 100\)
\(-16t^2 + 68t - 76 = 0\)
\(4t^2 - 17t + 19 = 0\)
b) Calculate Discriminant:
\(\Delta = 68^2 - 4(-16)(-76)\)
\(\Delta = 4624 - 4864\)
\(\Delta = -240 < 0\)
c) Safety Verdict:
Fails Safety Standard
Because \(\Delta < 0\), the mortar never reaches 100 feet (vertex peak is only \(96.25\text{ ft}\)).
18. [Consecutive Integers] Pneumatic Tosser Parameters
a) Solve for Integers:
\(n(n + 1) = 156\)
\(n^2 + n - 156 = 0\)
\(n = 12\text{ ft}\), \(n+1 = 13\text{ ft/s}\)
b) Height Equation:
\(h(t) = -16t^2 + v_0 t + h_0\)
\(h(t) = -16t^2 + 13t + 12\)
Apex: \(t \approx 0.41\text{ s}\)
c) Landing Time (\(h = 0\)):
\(t = \frac{-13 - \sqrt{169 + 768}}{-32}\)
\(t = \frac{-13 - \sqrt{937}}{-32}\)
\(t \approx 1.36\text{ seconds}\)
Review for Quiz • Vertex Form and Quadratic Applications • Teacher Answer Key Page 10 of 12
Step-by-step factoring of consecutive even integer dimensions and vertex form comparison.
Full-Page Solution
Given Trajectory: \(y = -\frac{1}{2}(x - 14)^2 + 98\) • Product of consecutive even base and height is 168.
Part A Solution: Set up Quadratic Equation and Factor
Let base \(= x\), and height \(= x + 2\) (consecutive positive even integers).
\(x(x + 2) = 168\)
\(x^2 + 2x - 168 = 0\)
Find two integers that multiply to \(-168\) and add to \(2\): \(\mathbf{14}\) and \(\mathbf{-12}\).
\((x + 14)(x - 12) = 0\)
Since dimensions must be positive, discard \(x = -14\). Thus \(x = 12\).
✓ Ramp Base = \(12\text{ feet}\), Ramp Height = \(12 + 2 = 14\text{ feet}\)
Part B Solution: Vertex Form Analysis & Altitude Clearance
Trajectory Function: \(y = -\frac{1}{2}(x - 14)^2 + 98\)
The function is given directly in vertex form \(y = a(x - h)^2 + k\), where \(a = -\frac{1}{2}\), \(h = 14\), and \(k = 98\).
The vertex is \((14, 98)\). Since \(a < 0\), the maximum vertical altitude is \(k = 98\text{ feet}\).
Compare Projectile Altitude to Barrier Height:
\(98\text{ ft (projectile max)} > 14\text{ ft (ramp height)}\) • Safely clears by \(84\text{ feet}\)!
Review for Quiz • Vertex Form and Quadratic Applications • Teacher Answer Key Page 11 of 12
Student formulation of vertical motion function followed by discriminant and airtime analysis.
Capstone Solution
Consecutive integer product \(n(n+1) = 56\) • Platform height \(h_0 = n\), Velocity \(v_0 = 10(n+1)\).
Part A Solution: Consecutive Integers & Height Model Setup
\(n(n + 1) = 56 \implies n^2 + n - 56 = 0 \implies (n + 8)(n - 7) = 0 \implies n = 7\)
Launch Height \(h_0 = 7\text{ feet}\), Initial Velocity \(v_0 = 10(7 + 1) = 80\text{ ft/s}\)
✓ Student Formulated Height Equation: \(h(t) = -16t^2 + 80t + 7\)
Part B Solution: Overhead Lighting Rig Clearance via Discriminant (\(h = 115\))
Set \(-16t^2 + 80t + 7 = 115 \implies -16t^2 + 80t - 108 = 0\)
\(\Delta = b^2 - 4ac = 80^2 - 4(-16)(-108) = 6400 - 6912 = \mathbf{-512 < 0}\)
Conclusion: Because \(\Delta < 0\), the equation has no real solutions. The performer cannot reach 115 feet. (Apex check: \(t = \frac{80}{32} = 2.5\text{ s} \implies h(2.5) = -16(6.25) + 80(2.5) + 7 = 107\text{ ft} < 115\text{ ft}\)).
Part C Solution: Quadratic Formula for Airtime to Net (\(h = 0\))
\(-16t^2 + 80t + 7 = 0\)
\(t = \frac{-80 \pm \sqrt{80^2 - 4(-16)(7)}}{-32} = \frac{-80 \pm \sqrt{6400 + 448}}{-32} = \frac{-80 \pm \sqrt{6848}}{-32}\)
\(\sqrt{6848} \approx 82.75 \implies t = \frac{-80 - 82.75}{-32} = \frac{-162.75}{-32} \approx 5.09\text{ seconds}\)
✓ Total Airtime \(\approx 5.09\text{ seconds}\)
Review for Quiz • Vertex Form and Quadratic Applications • Teacher Answer Key Page 12 of 12
