Math 133 Unit 2

Name Math133 Unit 2

Resolve the following mathematical problems: solving quadratic equations by factoring and using the quadratic formula; analyzing the graph of a quadratic function; working with a profit function; and plotting a parabola. Provide solutions with detailed work, explanations, and correctly formatted mathematical expressions.

Paper For Above instruction

In this assignment, we explore various fundamental algebraic concepts, including solving quadratic equations, graph analysis of quadratic functions, profit maximization, and graph plotting. Each section demonstrates specific techniques used in algebra and pre-calculus to develop problem-solving skills and mathematical understanding.

Problem 1: Solving Quadratic Equations

a) Solve the quadratic equation by factoring: x2 - 5x = 0

To factor the quadratic equation x2 - 5x = 0, first identify the greatest common factor (GCF) of the terms, which is x. Factoring out the GCF gives:

x(x - 5) = 0

Setting each factor equal to zero yields the solutions:

x = 0

x - 5 = 0 => x = 5

Therefore, the solutions are x = 0 and x = 5.

b) Solve the quadratic equation 3x2 + 2x - 16 = 0 using the quadratic formula

The quadratic formula is x = [-b ± √(b2 - 4ac)] / 2a

Given a = 3, b = 2, c = -16, substitute into the formula:

x = [-2 ± √(22 - 43(-16))] / (2*3)

x = [-2 ± √(4 + 192)] / 6

x = [-2 ± √196] / 6

x = [-2 ± 14] / 6

Calculate both solutions:

x = (-2 + 14) / 6 = 12 / 6 = 2

x = (-2 - 14) / 6 = -16 / 6 = -8/3

Solutions: x = 2 and x = -8/3.

Problem 2: Graph Analysis of y = x2 + 4x - 5

a) Determine the solutions of the equation x2 + 4x - 5 = 0 using only the graph.

Looking at the graph of y = x2 + 4x - 5, the solution(s) to the equation are the x-coordinates where the parabola crosses the x-axis. These intersection points are approximately at x ≈ -5 and x ≈ 1, based on the graph's intercepts.

Brief explanation: The x-intercepts of the parabola correspond to the solutions of the quadratic equation. By observing these points on the graph, we estimate the solutions without algebraic solving.

b) Does this function have a maximum or a minimum?

This parabola opens upward because the coefficient of x2 is positive (1). Therefore, the parabola has a minimum point at its vertex.

Explanation: Since the leading coefficient of the quadratic is positive, the parabola opens upwards, and the lowest point on the graph is a minimum.

c) What are the coordinates of the vertex in (x, y) form?

The vertex form of a quadratic is y = a(x - h)2 + k, where (h, k) are the vertex coordinates. Alternatively, find the vertex using -b / 2a for the x-coordinate:

xv = -4 / (2 * 1) = -4 / 2 = -2

Calculate y at x = -2:

y = (-2)2 + 4(-2) - 5 = 4 - 8 - 5 = -9

Vertex coordinates: (-2, -9)

d) What is the equation of the line of symmetry for this parabola?

The line of symmetry passes through the vertex and is given by x = h, where h is the x-coordinate of the vertex.

Line of symmetry: x = -2.

Problem 3: Profit Function Analysis

The profit function P(x) = -0.4x2 + fx - m, where f is the design fee, m is the monthly rent, and x is the number of awards designed per month.

a) If the design fee is $80, and the rent is $1,600, write the profit function P(x).

Substitute f = 80 and m = 1600 into the profit function:

P(x) = -0.4x2 + 80x - 1600

b) What is the profit when 50 awards are sold?

Calculate P(50):

P(50) = -0.4(50)2 + 80 * 50 - 1600

P(50) = -0.4(2500) + 4000 - 1600

P(50) = -1000 + 4000 - 1600 = 1400

Profit when 50 awards are designed: $1,400.

c) How many awards should be designed to maximize profit? Show your work algebraically.

This is a quadratic function with a negative leading coefficient, so its maximum occurs at the vertex x-coordinate:

xmax = -b / 2a = -80 / (2 * -0.4) = -80 / -0.8 = 100

Therefore, the profit is maximized when x = 100 awards.

d) What is the maximum profit?

Calculate P(100):

P(100) = -0.4(100)2 + 80 * 100 - 1600

P(100) = -0.4(10,000) + 8,000 - 1,600

P(100) = -4,000 + 8,000 - 1,600 = 2,400

Maximum profit achievable: $2,400.

Problem 4: Graphing a Parabola

a) Fill in the table with values of x and y for y = x2 - 6x

xy
-1(-1)^2 - 6(-1) = 1 + 6 = 7
00^2 - 6*0 = 0
11^2 - 6*1 = 1 - 6 = -5
24 - 12 = -8
39 - 18 = -9
416 - 24 = -8
525 - 30 = -5
636 - 36 = 0
749 - 42 = 7

b) Plot these points (x, y) on a graph to visualize the parabola. The line of symmetry can be found at the x-value of the vertex, which occurs at x = 3, the axis of symmetry.

To plot, use graphing tools or software like Excel, Desmos, or GeoGebra, and connect the points to see the shape of the parabola.

Conclusion

This assignment highlights essential algebraic skills, including solving quadratics by factoring and quadratic formula, analyzing parabola graphs, determining vertex properties, and optimizing quadratic profit functions. Mastery of these concepts enables deeper understanding of functions and their behaviors, forming a foundation for advanced mathematical study and applications in real-world scenarios.

References

  • Anton, H., Bivens, I., & Davis, S. (2016). Calculus: Early Transcendentals. Wiley.
  • Larson, R., & Edwards, B. H. (2017). Elementary Linear Algebra. Cengage Learning.
  • Lay, D. C. (2016). Linear Algebra and Its Applications. Pearson.
  • Stewart, J. (2015). Calculus: Concepts and Contexts. Cengage Learning.
  • Darrell, C. (2014). Precalculus with Limits: A Graphing Approach. Pearson.
  • Desmos. (n.d.). Graphing calculator. https://www.desmos.com/calculator
  • GeoGebra. (n.d.). Graphing calculator. https://www.geogebra.org/graphing
  • Khan Academy. (n.d.). Algebra and functions. https://www.khanacademy.org/math/algebra
  • MathWorld. (n.d.). Quadratic Equation. https://mathworld.wolfram.com/QuadraticEquation.html
  • Paul's Online Math Notes. (n.d.). Factoring quadratics. https://tutorial.math.lamar.edu/Classes/Alg/Factoring.asp

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