Exam Practice Notes

Exam Structure and Guidelines

• Exam Date: 2025
• Instructions:
  • To receive full credit, a correct answer must be thoroughly supported using proper mathematical notation.
  • Simplify answers when possible.
  • Calculators are prohibited on this exam.

Question 1: Polynomial Function

• Given function:
  $f(x) = x(x - 2)(x + 3)$
• Tasks: a. Sketch a graph of f.
  • Include all intercepts (x-intercepts and y-intercepts).
  • Exhibit proper end behavior.
    b. Solve the inequality: $f(x) ext{ is } ext{less than or equal to } 0.$
  • Write the answer using interval notation.

Question 2: Evaluation using Remainder Theorem

• Given polynomial:
  $g(x) = 3x^4 - 2x^3 + 4x^2 - 5x + 6 - 1$
• Task: Use the Remainder Theorem and synthetic division to find:
  $g(-1)$
• Answer space:
  $g(-1) = <strong><em><em></em></strong></em>$

Question 3: Properties of Polynomials

• Note: A polynomial has real coefficients, and one of the zeros is $2 + 3i$.
• Question: What must also be a zero of the polynomial?
  • Answer: The complex conjugate $2 - 3i$ must also be a zero.
  • Explanation: Since the coefficients of the polynomial are real, any non-real zeros occur in conjugate pairs.

Question 4: Analyzing Another Polynomial

• Given polynomial:
  $p(x) = 2x^4 - 5x^3 - 14x^2 + 47x - 30$
• Tasks: a. Use the rational zeros theorem to list all potential rational zeros for $p$. b. Find all zeros of $p$.
  • Answer space:
  • Zeros of p:_

Question 5: Function Analysis

• Given function:
  $r(x) = \frac{x^2 - x - 6}{x^2 - 9}$
• Tasks:
  a. State the equations of any horizontal asymptotes if there are any.
  b. Locate any holes if there are any.
  c. State the equations of any vertical asymptotes if there are any.
  d. State the domain of $r(x)$ in interval notation.
  e. Sketch a graph of $r$ illustrating the previous parts' information.

Question 6: Division of Polynomials

• Problem: Divide
  $(2x^3 - 5x^2 + 3x - 7) \div (x - 3)$
• Tasks: a. Find the quotient and remainder when performing the division:
  • Quotient: _
  • Remainder: _
    b. Explain how the quotient and remainder can assist in graphing $f(x)$.

Question 7: Profit Function Analysis

• Given Function:
  $P(x) = -2x^2 + 24x - 40$
• Tasks: a. Determine the x-intercepts and explain their significance in the context of this problem.
  • x-intercepts: $<strong><em><em></em></strong><strong><em></em></em></strong><strong><em>_</em></strong>$
    b. Find the x-value of the vertex and describe what this represents in the context of the problem.

Exam Structure and Guidelines

• Exam Date: 2025
• Instructions:
  • To receive full credit, a correct answer must be thoroughly supported using proper mathematical notation.
  • Simplify answers when possible.
  • Calculators are prohibited on this exam.

Question 1: Polynomial Function

• Given function: $f(x) = x(x - 2)(x + 3)$

• Tasks:
  a. Sketch a graph of f.

  • Include all intercepts (x-intercepts and y-intercepts).
  • Exhibit proper end behavior.
    Solution a:
  • X-intercepts: Set $f(x) = 0$, which gives $x = 0$, $x = 2$, $x = -3$. The x-intercepts are $(-3, 0)$, $(0, 0)$, $(2, 0)$.
  • Y-intercept: Set $x = 0$, which gives $f(0) = 0$. The y-intercept is $(0, 0)$.
  • End Behavior: The function is a cubic polynomial ($f(x) = x^3 + x^2 - 6x$) with a positive leading coefficient. Therefore, as $x \to -\infty$, $f(x) \to -\infty$ and as $x \to \infty$, $f(x) \to \infty$.
  • Sketch Description: The graph starts from the bottom left, passes through $(-3, 0)$, rises, turns, passes through $(0, 0)$, turns, passes through $(2, 0)$, and continues rising to the top right.

  b. Solve the inequality: $f(x) \text{ is less than or equal to } 0.$

  • Write the answer using interval notation.
    Solution b:
  • The critical points are the x-intercepts: $-3, 0, 2$.
  • Testing intervals:
    • For $x < -3$ (e.g., $x = -4$), $f(-4) = (-4)(-6)(-1) = -24 \le 0$.
    • For $-3 < x < 0$ (e.g., $x = -1$), $f(-1) = (-1)(-3)(2) = 6 > 0$.
    • For $0 < x < 2$ (e.g., $x = 1$), $f(1) = (1)(-1)(4) = -4 \le 0$.
    • For $x > 2$ (e.g., $x = 3$), $f(3) = (3)(1)(6) = 18 > 0$.
  • Thus, $f(x) \le 0$ on the intervals $(-\infty, -3]$ and $[0, 2]$.
  • Answer: $(-\infty, -3] \cup [0, 2]$

Question 2: Evaluation using Remainder Theorem

• Given polynomial: $g(x) = 3x^4 - 2x^3 + 4x^2 - 5x + 6 - 1$ which simplifies to $g(x) = 3x^4 - 2x^3 + 4x^2 - 5x + 5$
• Task: Use the Remainder Theorem and synthetic division to find: $g(-1)$
• Answer space:
  Solution:
  Using synthetic division with $-1$:

  -1 | 3  -2   4  -5   5
     |    -3   5  -9  14
     -----------------
       3  -5   9 -14  19

The remainder is $19$. By the Remainder Theorem, $g(-1)$ is the remainder.
$g(-1) = 19$

Question 3: Properties of Polynomials

• Note: A polynomial has real coefficients, and one of the zeros is $2 + 3i$.
• Question: What must also be a zero of the polynomial?
• Answer: The complex conjugate $2 - 3i$ must also be a zero.
• Explanation: Since the coefficients of the polynomial are real, any non-real zeros occur in conjugate pairs.

Question 4: Analyzing Another Polynomial

• Given polynomial: $p(x) = 2x^4 - 5x^3 - 14x^2 + 47x - 30$

• Tasks:
  a. Use the rational zeros theorem to list all potential rational zeros for $p$.
  Solution a:

  • Factors of the constant term $-30$ ($p$): $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$
  • Factors of the leading coefficient $2$ ($q$): $\pm 1, \pm 2$
  • Potential rational zeros $\frac{p}{q}$: $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{15}{2}$

  b. Find all zeros of $p$.

  • Answer space:
    Solution b:
  • Testing potential rational zeros:
    • For $x = 1$: $p(1) = 2(1)^4 - 5(1)^3 - 14(1)^2 + 47(1) - 30 = 2 - 5 - 14 + 47 - 30 = 0$. So, $x = 1$ is a zero.
  • Using synthetic division with $1$:

    1 | 2  -5  -14   47  -30
      |     2   -3  -17   30
      --------------------
        2  -3  -17   30    0
  

  The depressed polynomial is $2x^3 - 3x^2 - 17x + 30$.

  • Testing again (e.g., $x = -3$):
    • For $x = -3$: $2(-3)^3 - 3(-3)^2 - 17(-3) + 30 = -54 - 27 + 51 + 30 = 0$. So, $x = -3$ is a zero.
  • Using synthetic division with $-3$ on $2x^3 - 3x^2 - 17x + 30$:

    -3 | 2  -3  -17   30
       |    -6   27  -30
       -----------------
         2  -9   10    0
  

  The resulting quadratic is $2x^2 - 9x + 10 = 0$.

  • Solving the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. For $2x^2 - 9x + 10 = 0$, $a=2, b=-9, c=10$:
    • $x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(2)(10)}}{2(2)}$
    • $x = \frac{9 \pm \sqrt{81 - 80}}{4}$
    • $x = \frac{9 \pm \sqrt{1}}{4}$
    • $x = \frac{9 \pm 1}{4}$
    • This yields two zeros: $x = \frac{10}{4} = \frac{5}{2}$ and $x = \frac{8}{4} = 2$.
  • Zeros of $p$: $1, -3, \frac{5}{2}, 2$

Question 5: Function Analysis

• Given function: $r(x) = \frac{x^2 - x - 6}{x^2 - 9}$

• Tasks:
  a. State the equations of any horizontal asymptotes if there are any.
  Solution a:

  • First, simplify the function: $r(x) = \frac{(x - 3)(x + 2)}{(x - 3)(x + 3)}$. For $x \ne 3$, $r(x) = \frac{x + 2}{x + 3}$.
  • In the original function, the degree of the numerator (2) is equal to the degree of the denominator (2). The horizontal asymptote is $y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{1}{1} = 1$.
  • Equation of horizontal asymptote: $y = 1$

  b. Locate any holes if there are any.
  Solution b:

  • The factor $(x - 3)$ cancels out from the numerator and denominator, indicating a hole at $x = 3$.
  • To find the y-coordinate of the hole, substitute $x = 3$ into the simplified function $r(x) = \frac{x + 2}{x + 3}$: $y = \frac{3 + 2}{3 + 3} = \frac{5}{6}$.
  • Hole at $(3, \frac{5}{6})$

  c. State the equations of any vertical asymptotes if there are any.
  Solution c:

  • From the simplified function $r(x) = \frac{x + 2}{x + 3}$, the denominator becomes zero when $x + 3 = 0$, which means $x = -3$.
  • Equation of vertical asymptote: $x = -3$

  d. State the domain of $r(x)$ in interval notation.
  Solution d:

  • The denominator of the original function cannot be zero: $x^2 - 9 \ne 0 \implies (x - 3)(x + 3) \ne 0$.
  • Therefore, $x \ne 3$ and $x \ne -3$.
  • Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$

  e. Sketch a graph of $r$ illustrating the previous parts' information.
  Solution e:

  • The graph has a horizontal asymptote at $y = 1$, a vertical asymptote at $x = -3$, and a hole at $(3, \frac{5}{6})$. It passes through the x-intercept $(-2, 0)$ (from $x+2=0$ in the simplified function) and the y-intercept $(0, \frac{2}{3})$ (from $r(0) = \frac{0+2}{0+3}$). The curve approaches $y=1$ as $x \to \pm \infty$. As $x \to -3^-$, $r(x) \to \infty$; as $x \to -3^+$, $r(x) \to -\infty$.

Question 6: Division of Polynomials

• Problem: Divide $(2x^3 - 5x^2 + 3x - 7) \div (x - 3)$

• Tasks:
  a. Find the quotient and remainder when performing the division:

  • Quotient:
  • Remainder:
    Solution a:
    Using synthetic division with $3$:

  3 | 2  -5   3  -7
    |     6   3  18
    ----------------
      2   1   6  11
  

  • Quotient: $2x^2 + x + 6$
  • Remainder: $11$

  b. Explain how the quotient and remainder can assist in graphing $f(x)$.
  Solution b:

  • When a polynomial $f(x)$ is divided by $(x - c)$, the division results in $f(x) = q(x)(x - c) + R$, where $q(x)$ is the quotient and $R$ is the remainder.
  • According to the Remainder Theorem, $f(c) = R$. In this specific problem, dividing $f(x) = 2x^3 - 5x^2 + 3x - 7$ by $(x - 3)$ gives a remainder of $11$. This directly tells us that $f(3) = 11$.
  • Therefore, the quotient and remainder assist in graphing $f(x)$ by providing a specific point on the graph, which is $(3, 11)$.

Question 7: Profit Function Analysis

• Given Function: $P(x) = -2x^2 + 24x - 40$

• Tasks:
  a. Determine the x-intercepts and explain their significance in the context of this problem.

  • x-intercepts:
    Solution a:
  • To find the x-intercepts, set $P(x) = 0$:
    • $-2x^2 + 24x - 40 = 0$
    • Divide by $-2$: $x^2 - 12x + 20 = 0$
    • Factor the quadratic: $(x - 2)(x - 10) = 0$
  • X-intercepts: $x = 2$, $x = 10$
  • Significance: The x-intercepts represent the break-even points. These are the quantities (in terms of units, $x$) for which the profit $P(x)$ is zero. Producing or selling $2$ units or $10$ units results in no profit or loss.

  b. Find the x-value of the vertex and describe what this represents in the context of the problem.
  Solution b:

  • For a quadratic function in the form $Ax^2 + Bx + C$, the x-value of the vertex is given by the formula $x = -\frac{B}{2A}$.
  • For $P(x) = -2x^2 + 24x - 40$, we have $A = -2$ and $B = 24$.
  • $x = -\frac{24}{2(-2)} = -\frac{24}{-4} = 6$
  • Significance: The x-value of the vertex, $x = 6$, represents the number of units that must be produced and sold to maximize profit. At $$x=