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Quadratic Foundations Checkpoint

A free Algebra I lesson from the “Quadratic Foundations” unit, with a worked example and practice problems including step-by-step solutions.

This checkpoint samples all five Quadratic Foundations lessons: quadratic structure and factoring, solving by factoring and by completing the square, the quadratic formula and discriminant, graphs of quadratic functions, and quadratic applications. Questions are mixed, so start by naming the form you are given (factored, vertex, standard, or a context) and what is asked: a zero, a vertex, a number of solutions, or a measurement. Conventions: solutions and zeros mean real numbers; exact answers keep square roots (√); when a question asks you to round, enter the rounded number to the place it names; in a context, include only values that make sense, such as a positive length or a time after launch; and increasing or decreasing describes an interval of x-values. The zero-product property applies only when a product equals 0, and dividing by a variable can lose a solution. The 25 review questions and the 25 checkpoint questions were written separately from each other and from the lesson practice and quizzes. No question uses a diagram, and pairs of roots, exact radicals, and graph descriptions are multiple choice so that they are graded reliably. A score reports performance on this sample of proposed outcome A1.8; it is not an external certification. The written tasks below, including sketching a parabola, are reviewed with a rubric and do not change the score.

What you'll learn

Why it matters: Quadratics appear when two changing quantities are multiplied or when gravity pulls on a moving object: the area of a fenced garden, a frame or walkway around a picture or pool, the height of a thrown ball, and revenue when a higher price sells fewer tickets.

Worked example

Problem. Find the exact solutions of x² − 4x − 6 = 0, give them to the nearest hundredth, and describe where the graph of y = x² − 4x − 6 meets the x-axis.

  1. Identify a = 1, b = −4, and c = −6. The discriminant is (−4)² − 4(1)(−6) = 16 + 24 = 40, which is positive, so there are two real solutions.
  2. The quadratic formula gives x = (4 ± √40)/2. Because √40 = 2√10, this simplifies to x = 2 ± √10.
  3. Check by completing the square: x² − 4x = 6, so x² − 4x + 4 = 10, which is (x − 2)² = 10, and x − 2 = ±√10.
  4. √10 ≈ 3.162278, so the solutions are about 5.16 and −1.16. Round only at this last step.
  5. Because 40 is not a perfect square, the roots are irrational, so the trinomial has no factors with integer coefficients. The graph is a parabola opening upward with vertex (2, −10), crossing the x-axis at x ≈ 5.16 and x ≈ −1.16.

Answer: x = 2 + √10 or x = 2 − √10; about 5.16 and −1.16; the graph crosses the x-axis at those two x-values.

Lesson preview: this page shows 5 practice items. The interactive activity contains 25 practice items and 25 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.

Practice problems

1. Which of these functions is quadratic?

Choices: f(x) = (x + 4)(2 − x) · g(x) = x(x² − 9) · h(x) = (x + 3)² − x² · k(x) = 5x − 7

Show solution
  1. A quadratic function has degree 2 after it is written as a polynomial.
  2. Expanding gives f(x) = −x² − 2x + 8 (degree 2) and g(x) = x³ − 9x (degree 3).
  3. h(x) = x² + 6x + 9 − x² = 6x + 9: the squared terms cancel, so h is linear, as is k(x) = 5x − 7. Only f is quadratic.

Answer: f(x) = (x + 4)(2 − x)

2. The function g(x) = (2x − 5)(x + 3) has two zeros. What is the positive zero? Enter it as a decimal or a fraction.

Show solution
  1. A zero is an input that makes g(x) = 0, and a product is 0 exactly when one of its factors is 0.
  2. From 2x − 5 = 0, the value is x = 5/2 = 2.5; from x + 3 = 0, the value is x = −3.
  3. The positive zero is 2.5. Check: g(2.5) = (5 − 5)(5.5) = 0.

Answer: 2.5

3. Which plan correctly finds the zeros of f(x) = x² + 3x − 40?

Choices: Rewrite as (x + 8)(x − 5) = 0, then use x + 8 = 0 or x − 5 = 0 · Rewrite as (x − 8)(x + 5) = 0, then use x − 8 = 0 or x + 5 = 0 · Move 40 to get x² + 3x = 40, then take the square root of each side · Rewrite as x(x + 3) = 40, then use x = 40 or x + 3 = 40

Show solution
  1. Look for two numbers whose product is −40 and whose sum is 3: 8 and −5 work, so x² + 3x − 40 = (x + 8)(x − 5).
  2. The product equals 0 only when a factor is 0, so the zeros are −8 and 5. Check: 64 − 24 − 40 = 0 and 25 + 15 − 40 = 0.
  3. (x − 8)(x + 5) expands to x² − 3x − 40, a different polynomial. The square root of x² + 3x is not x + √3x, and x(x + 3) = 40 has a product equal to 40, not 0, so it cannot be split factor by factor.

Answer: Rewrite as (x + 8)(x − 5) = 0, then use x + 8 = 0 or x − 5 = 0

4. A bakery models its daily profit as P(n) = −(n − 4)(n − 22) hundred dollars, where n is the number of staff hours scheduled. What is the profit, in dollars, when n = 10?

Show solution
  1. Substitute n = 10: P(10) = −(10 − 4)(10 − 22) = −(6)(−12) = 72.
  2. P is measured in hundreds of dollars, so 72 hundred dollars is 72 × 100 = 7,200 dollars.
  3. The zeros 4 and 22 mark break-even staffing levels; 10 lies between them, so a positive profit makes sense.

Answer: 7200

5. For y = −2(x + 1)(x − 7), which statement is correct?

Choices: It opens downward, and its y-intercept is (0, 14) · It opens downward, and its y-intercept is (0, −14) · It opens upward, and its y-intercept is (0, 14) · It opens downward, and its y-intercept is (0, 7)

Show solution
  1. Expanded, the x² term is −2x², so the leading coefficient is −2 and the parabola opens downward.
  2. The y-intercept is the output at x = 0: y = −2(0 + 1)(0 − 7) = −2(1)(−7) = 14.
  3. So the graph meets the y-axis at (0, 14). Using only the factors gives (1)(−7) = −7 and drops the leading −2.

Answer: It opens downward, and its y-intercept is (0, 14)

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Choose and justify a method for each equation

Find all real solutions of each equation with a method you choose, and explain why that method fits: (a) x² − 2x − 24 = 0; (b) (x + 1)² = 12; (c) 2x² + 5x − 4 = 0. Give exact answers, then give (b) and (c) to the nearest hundredth. Check one solution of each equation in the original equation.

Review rubric and solution
  • (a) Factors as (x − 6)(x + 4) = 0 and uses the zero-product property to get x = 6 or x = −4, explaining that integer factors exist.
  • (b) Takes both square roots, x + 1 = ±√12 = ±2√3, to get x = −1 + 2√3 ≈ 2.46 or x = −1 − 2√3 ≈ −4.46, explaining that the equation is already a completed square.
  • (c) Uses the quadratic formula with discriminant 57 to get x = (−5 ± √57)/4 ≈ 0.64 or −3.14, explaining that 57 is not a perfect square, so integer factoring fails.
  • Keeps radicals exact until the last step and rounds to the nearest hundredth only when asked.
  • Shows a correct check of one solution for each equation in the original equation.
  1. (a) Two numbers with product −24 and sum −2 are −6 and 4, so (x − 6)(x + 4) = 0, and x = 6 or x = −4. Check: 36 − 12 − 24 = 0.
  2. (b) The left side is already a square, so x + 1 = √12 or x + 1 = −√12. Since √12 = 2√3, x = −1 + 2√3 ≈ 2.46 or x = −1 − 2√3 ≈ −4.46. Check: (2√3)² = 12.
  3. (c) a = 2, b = 5, c = −4, so b² − 4ac = 25 + 32 = 57. Then x = (−5 ± √57)/4. √57 ≈ 7.549834, so x ≈ 0.64 or x ≈ −3.14. Check: x ≈ 0.637459 gives 2x² + 5x − 4 ≈ 0.812707 + 3.187293 − 4 = 0, up to rounding.
  4. Factoring fits (a) because the discriminant 100 is a perfect square; square roots fit (b) because the square is already isolated; the formula fits (c) because its discriminant is not a perfect square.

Sketch a parabola and describe it on a restricted domain

For y = x² + 2x − 8: (a) write the function in vertex form and in factored form; (b) sketch its graph with labeled axes, the vertex, the axis of symmetry, both x-intercepts, the y-intercept, and one more pair of symmetric points other than the y-intercept and its mirror image; (c) state the x-intervals on which it increases and decreases; (d) state its range for all real x and its range when the domain is restricted to −3 ≤ x ≤ 1.

Review rubric and solution
  • Writes y = (x + 1)² − 9 and y = (x + 4)(x − 2), and shows that each expands back to x² + 2x − 8.
  • Labels the vertex (−1, −9), the axis x = −1, the x-intercepts (−4, 0) and (2, 0), the y-intercept (0, −8), and another symmetric pair such as (−3, −5) and (1, −5).
  • Draws a smooth upward parabola that is symmetric about x = −1 and passes through the labeled points.
  • States that the function decreases for x < −1 and increases for x > −1.
  • States the range y ≥ −9 for all real x and −9 ≤ y ≤ −5 for −3 ≤ x ≤ 1, explaining that the vertex lies inside the restricted domain.
  1. Completing the square: x² + 2x − 8 = (x² + 2x + 1) − 9 = (x + 1)² − 9. Factoring: two numbers with product −8 and sum 2 are 4 and −2, so y = (x + 4)(x − 2).
  2. The vertex is (−1, −9) and the axis is x = −1. The x-intercepts are (−4, 0) and (2, 0); the y-intercept is (0, −8), and its mirror across x = −1 is (−2, −8). Another symmetric pair is (−3, −5) and (1, −5), each 2 units from the axis.
  3. The parabola opens upward, so it decreases for x < −1 and increases for x > −1.
  4. For all real x the range is y ≥ −9. On −3 ≤ x ≤ 1, the vertex (−1, −9) is included and both endpoints give y = −5 (f(−3) = 9 − 6 − 8 = −5 and f(1) = 1 + 2 − 8 = −5), so the range is −9 ≤ y ≤ −5.

Model a throw and defend each answer in context

A ball is thrown upward, and its height is h(t) = −4.9t² + 14.7t + 1.2 meters, t seconds after the throw, until it lands. (a) Find the greatest height and when it occurs. (b) Find when the ball lands, to the nearest hundredth of a second, and explain why one root of h(t) = 0 is rejected. (c) State a reasonable domain and range for the model. (d) Explain which feature of the graph answers (a) and which answers (b). (e) Name one assumption of the model that a real throw might not meet.

Review rubric and solution
  • Finds the vertex time t = 14.7/9.8 = 1.5 seconds and the greatest height h(1.5) = 12.225 meters, with units.
  • Solves −4.9t² + 14.7t + 1.2 = 0 with the quadratic formula or another valid method to get t ≈ 3.08 seconds, and rejects t ≈ −0.08 because it is before the throw.
  • States the domain 0 ≤ t ≤ about 3.08 seconds and the range 0 ≤ h ≤ 12.225 meters, explaining both endpoints.
  • Explains that the vertex answers the greatest-height question while a zero answers the landing question.
  • Names a reasonable model assumption, such as no air resistance or a flat landing surface at height 0.
  1. The vertex time is t = −14.7/(2(−4.9)) = 1.5 seconds, and h(1.5) = −4.9(2.25) + 14.7(1.5) + 1.2 = −11.025 + 22.05 + 1.2 = 12.225 meters.
  2. Setting h(t) = 0 gives 4.9t² − 14.7t − 1.2 = 0, so t = (14.7 ± √(216.09 + 23.52))/9.8 = (14.7 ± √239.61)/9.8. √239.61 ≈ 15.479341, so t ≈ 3.08 or t ≈ −0.08. The negative time is before the throw, so the ball lands about 3.08 seconds after it is thrown.
  3. The model applies from the throw to the landing, so the domain is 0 ≤ t ≤ about 3.08 seconds, and the height runs from 0 m at landing up to the greatest height, so the range is 0 ≤ h ≤ 12.225 meters (the starting height, 1.2 m, lies inside this interval).
  4. The model assumes no air resistance and level ground at height 0; wind or landing on a slope would change the landing time.

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