Precalculus Exam Review: Polynomial, Rational, and Logarithmic Functions

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Q1. Sketch the graph of the given polynomial function. Discuss end behavior, zeros, cross/touch, p(x) < 0, p(x) > 0.

p(x) = -x^2 (x^2 + 2)(x + 4)
g(x) = -\frac{1}{5}(x + 5)(x - 4)^2
h(x) = \frac{1}{2}(x - 3)(x + 5)^3

Background

Topic: Polynomial Functions and Their Graphs

This question tests your understanding of how to analyze and sketch polynomial functions, including identifying zeros, their multiplicities, end behavior, and intervals where the function is positive or negative.

Key Terms and Formulas:

Zero: Value of x where p(x) = 0.
Multiplicity: The number of times a zero is repeated.
End Behavior: How the function behaves as x approaches ±∞.
Sign Analysis: Where p(x) < 0 or p(x) > 0.

Step-by-Step Guidance

Identify the degree and leading coefficient of each polynomial to determine end behavior. For example, for p(x), expand to see the highest power and its sign.
Find the zeros by setting each factor equal to zero. Note the multiplicity for repeated factors.
Determine whether the graph crosses or touches the x-axis at each zero (odd multiplicity = cross, even = touch).
Analyze intervals between zeros to determine where the function is positive or negative.
Set up a sign chart or use test points to check the sign of p(x) in each interval.

Try solving on your own before revealing the answer!

Final Answer:

Each polynomial's graph can be sketched by following the steps above. For example:

p(x) has zeros at x = 0 (multiplicity 2), x = -\sqrt{2}, x = \sqrt{2}, and x = -4. The end behavior is down on both sides due to the negative leading coefficient.
g(x) has zeros at x = -5 and x = 4 (multiplicity 2). The graph crosses at x = -5 and touches at x = 4.
h(x) has zeros at x = 3 and x = -5 (multiplicity 3). The graph crosses at both zeros, with a steeper slope at x = -5.

For each, use the sign chart and multiplicity to sketch the graph, showing intervals where p(x) < 0 and p(x) > 0.

Q2. Using the Rational Zeros Theorem, Descartes' Rule of Signs, and synthetic substitution, find the complex zeros of the given function. Write the function in complete factored form. Sketch the graph of p(x).

p(x) = 3x^4 + 5x^3 + 25x^2 + 45x - 18

Background

Topic: Finding Zeros of Polynomial Functions

This question tests your ability to use the Rational Zeros Theorem, Descartes' Rule of Signs, and synthetic substitution to find all zeros (real and complex) and write the polynomial in factored form.

Key Terms and Formulas:

Rational Zeros Theorem: Possible rational zeros are ± factors of the constant term divided by factors of the leading coefficient.
Descartes' Rule of Signs: Predicts the number of positive and negative real zeros.
Synthetic Substitution: A method to test possible zeros.

Step-by-Step Guidance

List all possible rational zeros using the Rational Zeros Theorem.
Apply Descartes' Rule of Signs to estimate the number of positive and negative real zeros.
Use synthetic substitution to test each possible rational zero.
Once a zero is found, use polynomial division to reduce the degree and repeat the process.
After finding all real zeros, use quadratic formula or factoring to find any complex zeros.

Try solving on your own before revealing the answer!

Final Answer:

The complete factored form is:

The zeros are x = -6, x = 1, and the solutions to (which are complex). The graph can be sketched using these zeros and the end behavior determined by the degree and leading coefficient.

Q3. Use the Intermediate Value Property to show there is at least one real zero between -1 and 0 for the given function.

p(x) = x^5 - 3x^2 + 4x + 2

Background

Topic: Intermediate Value Theorem

This question tests your understanding of how to use the Intermediate Value Theorem to prove the existence of a real zero in a given interval.

Key Terms and Formulas:

Intermediate Value Theorem: If a function is continuous on [a, b] and f(a) and f(b) have opposite signs, then there is at least one zero between a and b.

Step-by-Step Guidance

Evaluate p(-1) and p(0) to check their signs.
Confirm that p(x) is continuous on the interval [-1, 0].
Check if p(-1) and p(0) have opposite signs.
Apply the Intermediate Value Theorem to conclude there is at least one zero in the interval.

Try solving on your own before revealing the answer!

Final Answer:

p(-1) = -1 - 3 + (-4) + 2 = -6; p(0) = 2. Since p(-1) is negative and p(0) is positive, by the Intermediate Value Theorem, there is at least one real zero between -1 and 0.

Q4. Write a polynomial function, with real coefficients, that satisfies the given conditions. Write 2 versions: one with any "i", one without "i".

Degree 5; zeros: 4 (multiplicity 2), -2, 5i; p(3) = 10

Background

Topic: Constructing Polynomial Functions from Given Zeros

This question tests your ability to construct a polynomial given specific zeros, including complex zeros, and to adjust the leading coefficient to satisfy a value condition.

Key Terms and Formulas:

Multiplicity: The number of times a zero is repeated.
Complex zeros: If coefficients are real, complex zeros occur in conjugate pairs.
General form:

Step-by-Step Guidance

Write factors for each zero: (x - 4)^2, (x + 2), (x - 5i).
If coefficients must be real, include the conjugate zero: (x + 5i).
Multiply all factors to get the general form of the polynomial.
Set p(3) = 10 and solve for the leading coefficient a.

Try solving on your own before revealing the answer!

Final Answer:

With complex zero only:

With real coefficients:

To find a, substitute x = 3 and set p(3) = 10, then solve for a.

Q5. Sketch the graph of the given rational function. All intercepts, asymptotes, crossover point, and hole should be clearly indicated.

r(x) = \frac{3x^2 + 12x}{x^2 - x - 6}

Background

Topic: Rational Functions and Their Graphs

This question tests your ability to analyze and sketch rational functions, including finding intercepts, asymptotes, holes, and crossover points.

Key Terms and Formulas:

Vertical asymptote: Set denominator = 0 and solve for x.
Horizontal asymptote: Compare degrees of numerator and denominator.
Hole: Occurs if a factor cancels in numerator and denominator.
Intercepts: Set x = 0 for y-intercept; set numerator = 0 for x-intercepts.

Step-by-Step Guidance

Factor numerator and denominator to identify possible holes and vertical asymptotes.
Find x-intercepts by setting numerator = 0.
Find y-intercept by setting x = 0.
Determine vertical and horizontal asymptotes.
Check for holes by identifying common factors.

Try solving on your own before revealing the answer!

Final Answer:

Numerator factors: 3x(x + 4); Denominator factors: (x - 3)(x + 2). Vertical asymptotes at x = 3 and x = -2. No holes since no common factors. Horizontal asymptote at y = 3 (degrees equal). x-intercepts at x = 0 and x = -4. y-intercept at r(0) = 0. Sketch the graph with these features.

Q6. Find the domain of the given function.

f(x) = log(4 - x^2)
f(x) = \ln(x^2)
f(x) = log(x^3)

Background

Topic: Domain of Logarithmic Functions

This question tests your ability to find the domain of logarithmic functions, which depends on the argument being positive.

Key Terms and Formulas:

Domain: Set argument of log or ln > 0.
log(x): Defined for x > 0.
ln(x): Defined for x > 0.

Step-by-Step Guidance

For each function, set the argument > 0 and solve for x.
For log(4 - x^2), solve 4 - x^2 > 0.
For ln(x^2), solve x^2 > 0.
For log(x^3), solve x^3 > 0.

Try solving on your own before revealing the answer!

Final Answer:

f(x) = log(4 - x^2): Domain is -2 < x < 2
f(x) = ln(x^2): Domain is x ≠ 0 (all real numbers except 0)
f(x) = log(x^3): Domain is x > 0

Q7. Sketch the graph of the given function: f(x) = -log(9 - x) - 2

Background

Topic: Transformations of Logarithmic Functions

This question tests your ability to graph logarithmic functions with transformations, including shifts and reflections.

Key Terms and Formulas:

Vertical shift: f(x) + c shifts graph up/down.
Reflection: Negative sign reflects graph across x-axis.
Domain: Argument of log must be positive.

Step-by-Step Guidance

Identify the base function: log(9 - x).
Determine the domain: 9 - x > 0 ⇒ x < 9.
Apply the reflection: Multiply output by -1.
Apply the vertical shift: Subtract 2 from the output.
Sketch the graph using these transformations.

Try solving on your own before revealing the answer!

Final Answer:

The graph is a reflected and shifted logarithmic curve, defined for x < 9, with a vertical asymptote at x = 9 and shifted down by 2 units.