MAT 105 College Algebra - Final Exam Review Notes

Simplify Expressions

• a. Simplifying Exponential Expressions:
  (-5x^2) \cdot (-7x^3) = 35x^5
• b. Power of a Power:
  (2z^3)^4 = 16z^{12}
• c. Dividing Exponential Expressions:
  \frac{10y^7}{5y^3} = 2y^4

Simplify Radicals

• a. Simplifying Radicals:
  \sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}
• b. Simplifying Negative Radicals:
  -\sqrt{72} = -\sqrt{36 \cdot 2} = -6\sqrt{2}

Solve Linear Equations

• Solving for u:
  -5(u - 4) + 2 = 11 - (u - 3)
  -5u + 20 + 2 = 11 - u + 3
  -5u + 22 = 14 - u
  -4u = -8
  u = 2

Solve for a Variable

• Solving for r:
  q = \frac{c}{4(h + r)}
  4q(h + r) = c
  h + r = \frac{c}{4q}
  r = \frac{c}{4q} - h

Simplify Complex Numbers

• Simplifying Products of Imaginary Numbers:
  \sqrt{-25} \sqrt{-81} = (5i)(9i) = 45i^2 = -45

Perform Operations with Complex Numbers

• Multiplying Complex Numbers:
  (6 + 5i)(-9 + 9i) = -54 + 54i - 45i + 45i^2 = -54 + 9i - 45 = -99 + 9i

Perform Operations with Complex Numbers - Division

• Dividing Complex Numbers:
  \frac{-8 + 3i}{5 + 7i} = \frac{(-8 + 3i)(5 - 7i)}{(5 + 7i)(5 - 7i)} = \frac{-40 + 56i + 15i - 21i^2}{25 - 49i^2} = \frac{-40 + 71i + 21}{25 + 49} = \frac{-19 + 71i}{74} = -\frac{19}{74} + \frac{71}{74}i

Solve Quadratic Equations

• Solving Quadratic Equations by Factoring:
  x^2 + 2x - 120 = 0
  (x + 12)(x - 10) = 0
  x = -12, 10

Use Quadratic Formula

• Using the Quadratic Formula:
  3x^2 - 13 = 0
  x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{0 \pm \sqrt{0^2 - 4(3)(-13)}}{2(3)} = \frac{\pm \sqrt{156}}{6} = \frac{\pm 2\sqrt{39}}{6} = \pm \frac{\sqrt{39}}{3}

Solve Radical Equations

• Solving Radical Equations:
  \sqrt{4x + 5} = 9
  4x + 5 = 81
  4x = 76
  x = 19

Solve Absolute Value Equations

• Solving Absolute Value Equations:
  |6z - 3| = 7
  6z - 3 = 7 \quad \text{or} \quad 6z - 3 = -7
  6z = 10 \quad \text{or} \quad 6z = -4
  z = \frac{5}{3}, -\frac{2}{3}

Solve Equations with Substitution

• Solving Equations Using Substitution:
  (t + 3)^2 - (t + 3) - 12 = 0
  Let u = t + 3
  u^2 - u - 12 = 0
  (u - 4)(u + 3) = 0
  u = 4, -3
  t + 3 = 4 \quad \text{or} \quad t + 3 = -3
  t = 1, -6

Solve Compound Inequalities

• Solving Compound Inequalities:
  -6 \le -3x + 9 < 0 -15 \le -3x < -9 5 \ge x > 3
  Interval Notation: (3, 5]

Solve Linear Inequalities

• Solving Linear Inequalities:
  9(x - 3) - 8x \ge -3
  9x - 27 - 8x \ge -3
  x \ge 24
  Interval Notation: [24, \infty)

Solve Absolute Value Inequalities (Less Than)

• Solving Absolute Value Inequalities (Less Than):
  |x + 8| + 3 \le 6
  |x + 8| \le 3
  -3 \le x + 8 \le 3
  -11 \le x \le -5
  Interval Notation: [-11, -5]

Solve Absolute Value Inequalities (Greater Than)

• Solving Absolute Value Inequalities (Greater Than):
  |w - 3| > 4
  w - 3 > 4 \quad \text{or} \quad w - 3 < -4 w > 7 \quad \text{or} \quad w < -1
  Interval Notation: (-\infty, -1) \cup (7, \infty)

Distance and Midpoint

• a. Distance Between Two Points:
  Points: (3, 6) and (-4, -1)
  d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} = \sqrt{(-4 - 3)^2 + (-1 - 6)^2} = \sqrt{(-7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2}
• b. Midpoint of a Line Segment:
  M = (\frac{x1 + x2}{2}, \frac{y1 + y2}{2}) = (\frac{3 + (-4)}{2}, \frac{6 + (-1)}{2}) = (-\frac{1}{2}, \frac{5}{2})

Equation of a Circle

• Determining Center and Radius:
  (x - 7)^2 + (y - 3)^2 = 25
  Center: (7, 3), Radius: 5

Equation of a Circle from Center and Radius

• Equation of a Circle in Standard Form:
  Center: (-2, 5), Radius: 1
  (x + 2)^2 + (y - 5)^2 = 1

Completing the Square

• Completing the Square and Finding Center and Radius:
  x^2 + 18x + 81 + y^2 - 8y + 16 = 81
  (x + 9)^2 + (y - 4)^2 = 81
  Center: (-9, 4), Radius: 9

Relations and Functions

• a. Set of Ordered Pairs:
  Given relation.
• b. Domain of the Relation:
  The set of all x-values in the relation: \text{Domain} = {-4, -1, 0, 2}
• c. Range of the Relation:
  The set of all y-values in the relation: \text{Range} = {-3, -2, 2, 3, 4}
• d. Function Determination:
  The relation is not a function because the x-value -1 is associated with two different y-values (2 and 4).

Intercepts of a Function

• Finding Intercepts:
  k(x) = -|x| + 2
  x-intercepts: (-2, 0), (2, 0)
  y-intercept: (0, 2)

Slope of a Line

• Finding the Slope:
  Points: (-2, -6) and (-9, -17)
  m = \frac{y2 - y1}{x2 - x1} = \frac{-17 - (-6)}{-9 - (-2)} = \frac{-11}{-7} = \frac{11}{7}

Average Rate of Change

• Average Rate of Change:
  f(x) = -2x^2 - 3 from x = -2 to x = 1
  f(-2) = -2(-2)^2 - 3 = -11
  f(1) = -2(1)^2 - 3 = -5
  \text{Average Rate of Change} = \frac{f(1) - f(-2)}{1 - (-2)} = \frac{-5 - (-11)}{3} = \frac{6}{3} = 2

Equation of a Parallel Line

• Equation of a Line Parallel to Another Line:
  Point: (2, 3), Parallel to: y = -2x + 3
  y - y1 = m(x - x1)
  y - 3 = -2(x - 2)
  y = -2x + 7

Equation of a Perpendicular Line

• Equation of a Line Perpendicular to Another Line:
  Point: (5, 2), Perpendicular to: 3y - x = 9 \Rightarrow y = \frac{1}{3}x + 3
  Perpendicular slope: -3
  y - 2 = -3(x - 5)
  y = -3x + 17

Transformations of Square Root Functions

• Graphing Transformations:
  g(x) = -\sqrt{x + 2} - 1
  Start with f(x) = \sqrt{x}.

Graph Transformations

• Graph Transformations:
  y = -f(x - 2) + 4
  Given graph of y = f(x).

Even, Odd, or Neither

• Determining Even, Odd, or Neither:
  f(x) = x^5 - x^4
  f(-x) = (-x)^5 - (-x)^4 = -x^5 - x^4
  Since f(-x) \neq f(x) and f(-x) \neq -f(x), the function is neither even nor odd.

Piecewise Functions

• Evaluating Piecewise Functions: g(x) = \begin{cases} -2|x| - 3 & \text{for } x \le -2 \ 5x + 6 & \text{for } -2 < x < 3 \ 4 & \text{for } x \ge 3 \end{cases}
  • a. g(-3) = -2|-3| - 3 = -2(3) - 3 = -9
  • b. g(3) = 4
  • c. g(-2) = -2|-2| - 3 = -2(2) - 3 = -7
  • d. g(0) = 5(0) + 6 = 6
  • e. g(4) = 4

Analyzing Function Graphs

• Analyzing Graphs of Functions: Given graph of y = f(x).
  • a. f(2) = -1
  • b. Domain: (-4, \infty)
  • c. Range: [-2, \infty)
  • d. Increasing: (1, \infty)
  • e. Decreasing: (-1, 1)
  • f. Constant: (-4, -1)

Algebra of Functions

• Algebra of Functions: f(x) = x - 4, \quad g(x) = \frac{1}{x - 3}, \quad h(x) = \sqrt{x - 5}
  • a. (f - h)(6) = f(6) - h(6) = (6 - 4) - \sqrt{6 - 5} = 2 - 1 = 1
  • b. (g \cdot h)(5) = g(5) \cdot h(5) = \frac{1}{5 - 3} \cdot \sqrt{5 - 5} = \frac{1}{2} \cdot 0 = 0
  • c. (h \circ f)(1) = h(f(1)) = h(1 - 4) = h(-3) = \sqrt{-3 - 5} = \sqrt{-8}, which is undefined.
  • d. (f \cdot g)(x) = f(x) \cdot g(x) = (x - 4) \cdot \frac{1}{x - 3} = \frac{x - 4}{x - 3}

Composition of Functions

• Composition of Functions:
  f(x) = 3x + 9, \quad g(x) = 5x - 1
  (f \circ g)(x) = f(g(x)) = f(5x - 1) = 3(5x - 1) + 9 = 15x - 3 + 9 = 15x + 6

Finding f(x+h)

• Finding and Simplifying Expressions: f(x) = 8x + 4
  • a. f(x + h) = 8(x + h) + 4 = 8x + 8h + 4
  • b. \frac{f(x + h) - f(x)}{h} = \frac{(8x + 8h + 4) - (8x + 4)}{h} = \frac{8h}{h} = 8

Quadratic Functions

• Finding Quadratic Functions:
  Vertex: (-3, 1), Passes through: (0, -17)
  f(x) = a(x - h)^2 + k
  f(x) = a(x + 3)^2 + 1
  -17 = a(0 + 3)^2 + 1
  -18 = 9a
  a = -2
  f(x) = -2(x + 3)^2 + 1

Analyzing Quadratic Functions

• Identifying Vertex, Axis of Symmetry, and Intercepts:
  f(x) = -3x^2 + 12x - 9
  Vertex: (2, 3), Axis of Symmetry: x = 2, x-intercepts: (1, 0), (3, 0), y-intercept: (0, -9)

Leading Coefficient Test

• Leading Coefficient Test:
  • a. f(x) = 4x^4 + 4x^3 + 4x^2 + 5x + 3
    As x \to -\infty, f(x) \to \infty and as x \to \infty, f(x) \to \infty
  • b. g(x) = -3x^3 - 2x^2 + 3x + 3
    As x \to -\infty, g(x) \to \infty and as x \to \infty, g(x) \to -\infty

Zeros of Polynomials

• Finding Zeros, Multiplicities, and Cross/Touch Points:
  f(x) = 3(x + 2)(x + 5)^2
  Zero: -2, Multiplicity: 1, Cross point.
  Zero: -5, Multiplicity: 2, Touch point.

Remainder Theorem

• Using the Remainder Theorem:
  f(x) = x^3 + 4x^2 + 3x - 5, Evaluate f(-3)
  f(-3) = (-3)^3 + 4(-3)^2 + 3(-3) - 5 = -27 + 36 - 9 - 5 = -5

Polynomial Division

• Polynomial Division and Synthetic Division:
  Dividend: x^3 - 2x^2 - 25x - 4
  Divisor: x + 4
  Quotient: x^2 - 6x - 1
  Remainder: 0

Constructing Polynomials

• Constructing a Polynomial from Zeros:
  Zeros: 4, 6, -1, Leading coefficient: 1
  f(x) = (x - 4)(x - 6)(x + 1) = (x^2 - 10x + 24)(x + 1) = x^3 - 9x^2 + 14x + 24

Constructing Polynomials with Complex Zeros

• Polynomial from Zeros (Including Complex):
  Zeros: -2, 3i, -3i, Leading coefficient: 1
  f(x) = (x + 2)(x - 3i)(x + 3i) = (x + 2)(x^2 + 9) = x^3 + 2x^2 + 9x + 18

Possible Rational Zeros

• Listing Possible Rational Zeros:
  f(x) = 6x^4 + 2x^3 - 3x^2 + 2
  \pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{1}{6}

Vertical Asymptotes

• Identifying Vertical Asymptotes:
  f(x) = \frac{8}{x - 4}
  Vertical Asymptote: x = 4

Vertical Asymptotes of Rational Functions

• Identifying Vertical Asymptotes:
  g(x) = \frac{8x^2 - 6}{x^2 - 6x + 5} = \frac{8x^2 - 6}{(x - 1)(x - 5)}
  Vertical Asymptotes: x = 1, x = 5

Vertical Asymptotes - None

• Identifying Vertical Asymptotes:
  f(x) = \frac{x^3 - 6}{x^2 + 25}
  No vertical asymptote because x^2 + 25 has no real roots.

Inverse Functions

• Determining if Functions are Inverses:
  g(x) = \frac{x}{6} + 2, \quad h(x) = 6x - 2
  g(h(x)) = \frac{6x - 2}{6} + 2 = x - \frac{1}{3} + 2 = x + \frac{5}{3} \neq x
  Not inverses.

Horizontal Asymptotes

• Identifying Horizontal Asymptotes:
  f(x) = \frac{8}{x - 4}
  Horizontal Asymptote: y = 0

Inverse of a Function

• Finding the Inverse of a Function:
  f(x) = 3x - 7
  y = 3x - 7
  x = 3y - 7
  3y = x + 7
  y = \frac{x + 7}{3}
  f^{-1}(x) = \frac{x + 7}{3}

Inverse of a One-to-One Function

• Finding the Inverse of a One-to-One Function:
  k(x) = \sqrt{x + 7}
  y = \sqrt{x + 7}
  x = \sqrt{y + 7}
  x^2 = y + 7
  y = x^2 - 7
  k^{-1}(x) = x^2 - 7