Axis of Sym. \(x = -\frac{3}{2}\)
Vertex \((-\frac{3}{2}, -\frac{19}{2})\)
\(y\)-Intercept \((0, -5)\)
Algebra 2 • Parabola Features Unit Answer Key Page 1 of 2
Part II Solutions
Teacher Reference Key
Full algebraic substitutions and contextual interpretations for multi-form analysis.
5 \(y = -3(x - 4)^2 + 12\) Vertex Form
Vertex: Directly from \(y = a(x - h)^2 + k \implies (4, 12)\).
AOS: Vertical line through vertex \(\implies x = 4\).
\(y\)-int: Set \(x=0 \implies y = -3(0 - 4)^2 + 12 = -3(16) + 12 = -48 + 12 = \mathbf{-36}\).
AOS \(x = 4\)
Vertex \((4, 12)\)
\(y\)-Int \((0, -36)\)
6 \(f(x) = \frac{1}{2}(x + 6)^2 - 8\) Vertex Form
Vertex: Note \((x+6) = (x - (-6)) \implies (-6, -8)\).
AOS: \(x = -6\).
\(y\)-int: Set \(x=0 \implies f(0) = \frac{1}{2}(6)^2 - 8 = \frac{1}{2}(36) - 8 = 18 - 8 = \mathbf{10}\).
AOS \(x = -6\)
Vertex \((-6, -8)\)
\(y\)-Int \((0, 10)\)
7 \(g(x) = -(x - 1)(x + 7)\) Factored Form
Roots: \(x = 1\) and \(x = -7\).
AOS (Midpoint): \(x = \frac{1 + (-7)}{2} = \frac{-6}{2} = \mathbf{-3}\).
Vertex: \(g(-3) = -(-4)(4) = -(-16) = \mathbf{16}\).
\(y\)-int: \(g(0) = -(-1)(7) = \mathbf{7}\).
AOS \(x = -3\)
Vertex \((-3, 16)\)
\(y\)-Int \((0, 7)\)
8 Honors
\(h(x) = -\frac{1}{4}(x - 2)(x - 10)\) Factored
Roots: \(x = 2,\; x = 10\).
AOS: \(x = \frac{2 + 10}{2} = \mathbf{6}\).
Vertex: \(h(6) = -\frac{1}{4}(4)(-4) = -\frac{1}{4}(-16) = \mathbf{4}\).
\(y\)-int: \(h(0) = -\frac{1}{4}(-2)(-10) = -\frac{20}{4} = \mathbf{-5}\).
AOS \(x = 6\)
Vertex \((6, 4)\)
\(y\)-Int \((0, -5)\)
9 Honors Challenge
Parameters
1. Max value 6 at AOS \(x=3 \implies\) Vertex is \((3, 6)\).
2. AOS: \(-\frac{b}{2a} = 3 \implies b = -6a\).
3. Plug into \(y = ax^2 + bx - 12\):
\(6 = a(9) + (-6a)(3) - 12 \implies 6 = -9a - 12 \implies -9a = 18 \implies \mathbf{a = -2}\).
4. \(b = -6(-2) = \mathbf{12}\). Function: \(y = -2x^2 + 12x - 12\).
\(a\) and \(b\) \(a=-2, b=12\)
Vertex \((3, 6)\)
\(y\)-Int \((0, -12)\)
10 Honors Modeling
Contextual
AOS / Peak Time: \(t = -\frac{64}{2(-16)} = \frac{-64}{-32} = \mathbf{2\text{ s}}\).
Vertex / Max Height: \(h(2) = -16(4) + 64(2) + 80 = -64 + 128 + 80 = \mathbf{144\text{ ft}}\). Vertex: \((2, 144)\).
Vertical Intercept: \(h(0) = \mathbf{80\text{ ft}}\). Physical meaning: rocket launched from a platform 80 ft high.
Peak Time \(t = 2\text{ s}\)
Max Height \(h = 144\text{ ft}\)
Launch Elev. \((0, 80\text{ ft})\)
Algebra 2 • Parabola Features Unit Answer Key Page 2 of 2
