Quadratic functions are one of the most important topics in Algebra 2. They connect directly to the Algebra 1 work students did with factoring and linear equations, while introducing new tools — vertex form, the discriminant, and complex roots — that appear throughout precalculus and calculus.
This post provides practice problems across the key quadratic function skills tested in Algebra 2 courses, end-of-course exams, and college placement tests, along with full worked solutions.
Algebra 2 Quadratic Functions: Key Topics
Before diving into practice problems, here is a quick reference for the core concepts in this unit:
- Standard form: f(x) = ax² + bx + c. The vertex x-coordinate is x = −b/(2a).
- Vertex form: f(x) = a(x − h)² + k. The vertex is (h, k). The parabola opens up if a > 0 and down if a < 0.
- Factored form: f(x) = a(x − r₁)(x − r₂). The roots (zeros) are x = r₁ and x = r₂.
- Quadratic formula: x = [−b ± √(b² − 4ac)] / 2a
- Discriminant: b² − 4ac. Positive → two real roots. Zero → one real root. Negative → two complex (non-real) roots.
Practice Problem Set 1: Graphing and Vertex
Problem 1. Identify the vertex and axis of symmetry for f(x) = 2x² − 8x + 3.
Solution: Use x = −b/(2a) = −(−8)/[2(2)] = 8/4 = 2. Then f(2) = 2(4) − 8(2) + 3 = 8 − 16 + 3 = −5. Vertex: (2, −5). Axis of symmetry: x = 2.
Problem 2. Write f(x) = x² − 6x + 11 in vertex form.
Solution: Complete the square. f(x) = (x² − 6x + 9) + 2 = (x − 3)² + 2. Vertex form: f(x) = (x − 3)² + 2. Vertex: (3, 2).
Problem 3. A parabola has vertex (−1, 4) and passes through (1, −4). Write the equation in vertex form.
Solution: f(x) = a(x + 1)² + 4. Substitute (1, −4): −4 = a(4) + 4 → −8 = 4a → a = −2. Answer: f(x) = −2(x + 1)² + 4.
Practice Problem Set 2: Finding Zeros
Problem 4. Solve by factoring: x² − 7x + 12 = 0.
Solution: Factor as (x − 3)(x − 4) = 0. Solutions: x = 3 and x = 4.
Problem 5. Solve using the quadratic formula: 2x² + 3x − 5 = 0.
Solution: a = 2, b = 3, c = −5. Discriminant = 9 + 40 = 49. x = (−3 ± 7)/4. Solutions: x = 1 and x = −5/2.
Problem 6. How many real solutions does 3x² − 4x + 5 = 0 have?
Solution: Discriminant = (−4)² − 4(3)(5) = 16 − 60 = −44. Since the discriminant is negative, there are no real solutions (two complex conjugate solutions).
Problem 7. Solve by completing the square: x² + 10x − 3 = 0.
Solution: x² + 10x = 3. Add 25 to both sides: (x + 5)² = 28. x + 5 = ±√28 = ±2√7. Solutions: x = −5 ± 2√7.
Practice Problem Set 3: Real-World Applications
Problem 8. A ball is thrown upward from the ground with an initial velocity of 64 ft/sec. Its height is modeled by h(t) = −16t² + 64t. At what time does the ball reach its maximum height, and what is that height?
Solution: Vertex x-coordinate: t = −64/[2(−16)] = 2 seconds. h(2) = −16(4) + 64(2) = −64 + 128 = 64 feet. Maximum height is 64 feet at t = 2 seconds.
Problem 9. A rectangular garden has a perimeter of 60 meters. Express the area as a function of the width w, and find the width that maximizes the area.
Solution: Length = 30 − w. Area = w(30 − w) = −w² + 30w. Vertex: w = −30/[2(−1)] = 15. Maximum area at width = 15 m (a square garden).
Problem 10. The revenue from selling x items is R(x) = −2x² + 120x. What number of items maximizes revenue, and what is the maximum revenue?
Solution: Vertex: x = −120/[2(−2)] = 30 items. R(30) = −2(900) + 120(30) = −1800 + 3600 = $1,800. Sell 30 items for maximum revenue of $1,800.
Practice Problem Set 4: Quadratic Inequalities
Problem 11. Solve the inequality: x² − 5x + 6 < 0.
Solution: Factor as (x − 2)(x − 3) < 0. The parabola opens up and equals zero at x = 2 and x = 3. The quadratic is negative between the roots. Solution: 2 < x < 3.
Problem 12. Solve: x² + x − 12 ≥ 0.
Solution: Factor as (x + 4)(x − 3) ≥ 0. Roots at x = −4 and x = 3. Parabola opens up; it is non-negative outside the roots. Solution: x ≤ −4 or x ≥ 3.
Common Algebra 2 Quadratic Mistakes
- Vertex form sign confusion: In f(x) = (x − h)² + k, the vertex is at (h, k) — not (−h, k). Students regularly reverse the sign of h.
- Completing the square with a leading coefficient ≠ 1: Always factor out the leading coefficient before completing the square, or you will get the wrong constant term to add.
- Forgetting ± in the quadratic formula: Every square root step produces two answers. Write ± explicitly every time.
- Interpreting the discriminant: A discriminant of zero means exactly one real root (the vertex touches the x-axis), not “no solution.” Only a negative discriminant means no real solutions.
Algebra 2 Practice Resources from ViewMath
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