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Quizzes > High School Quizzes > Mathematics

Pre Calculus Unit 3 Practice Quiz

Ace Unit 3 with clear test answers

Difficulty: Moderate
Grade: Grade 12
Study OutcomesCheat Sheet
Paper art promoting a trivia quiz on Precalc Unit 3 for high school students.

What is the domain of the function f(x) = √x?
x ≥ 0
x > 0
All real numbers
x ≤ 0
The square root function is only defined for nonnegative arguments, so the domain is x ≥ 0. The other options fail to capture the requirement for nonnegative x values.
What is the effect of the transformation f(x) = g(x) + 3 on the graph of g(x)?
Vertical shift upward by 3
Horizontal shift to the right by 3
Vertical stretch by a factor of 3
Reflection across the x-axis
Adding 3 to the function values shifts the graph vertically upward by 3. This transformation does not affect the horizontal alignment or scale of the graph.
If f(x) = 2x and g(x) = x + 5, what is (f ∘ g)(x)?
2x + 10
2x + 5
x + 10
2x + 7
The composition (f ∘ g)(x) is computed by substituting g(x) into f, resulting in f(g(x)) = 2*(x + 5) = 2x + 10. The other options do not correctly follow the order of operations required for composition.
What is the inverse function of f(x) = x + 2?
f❻¹(x) = x - 2
f❻¹(x) = 2 - x
f❻¹(x) = -x + 2
f❻¹(x) = 1/(x + 2)
To find the inverse function, you swap x and y in the equation y = x + 2 and solve for y, which gives y = x - 2. This confirms that f❻¹(x) = x - 2 is the correct inverse.
Which of the following is a linear function?
f(x) = 3x - 4
f(x) = x²
f(x) = √x
f(x) = 1/x
A linear function has the form f(x) = mx + b. The function 3x - 4 fits this form while the other options represent quadratic, radical, and rational functions.
What is the y-intercept of the quadratic function f(x) = 2x² - 3x + 1?
1
2
-3
0
The y-intercept is found by evaluating the function at x = 0. For f(x) = 2x² - 3x + 1, f(0) equals 1, so the y-intercept is 1.
Solve for x in the equation 3^x = 27.
3
2
4
1
Since 27 can be written as 3³, equating 3^x to 3³ shows that x must equal 3. This makes 3 the unique solution based on properties of exponents.
For a quadratic function f(x) = ax² + bx + c, what is the formula to find the x-coordinate of its vertex?
-b/(2a)
b/(2a)
-2a/b
2a/b
The x-coordinate of the vertex of a quadratic function is given by -b/(2a). This formula is derived from either completing the square or using calculus to find the function's extreme value.
What transformation does the function h(x) = -f(x) represent?
Reflection over the x-axis
Reflection over the y-axis
Vertical shift downward
Horizontal shift left
Multiplying a function by -1 reflects its graph over the x-axis, changing the sign of all its y-values. This is a standard transformation in function analysis.
If f(x) = x² and g(x) = x + 1, what is (f ∘ g)(x)?
(x + 1)²
x² + 1
x² + 2
x² - 1
The composition (f ∘ g)(x) is obtained by substituting g(x) into f(x), which gives (x + 1)². This correctly represents the composite function before any expansion.
What is the domain of the logarithmic function f(x) = log(x - 2)?
x > 2
x ≥ 2
x < 2
All real numbers
For the logarithmic function log(x - 2) to be defined, the argument x - 2 must be positive. Therefore, x must be greater than 2.
Which of the following represents an exponential growth function?
f(x) = 3 · 2^x
f(x) = 3 · (1/2)^x
f(x) = -2^x
f(x) = log(x)
An exponential growth function features a base greater than 1; here, 2^x grows exponentially as x increases. Multiplying by 3 does not change the growth behavior.
Find the domain of the function f(x) = 1/(x² - 9).
All real numbers except x = 3 and x = -3
All real numbers
x > 3 or x < -3
x = 3 or x = -3
The function is undefined when the denominator is zero. Since x² - 9 factors into (x - 3)(x + 3), x cannot be 3 or -3.
Which expression represents the composition (f ∘ g)(x) if f(x) = x² and g(x) = √x?
(√x)²
x^(3/2)
√(x²)
x√x
The composition (f ∘ g)(x) is f(g(x)) which equals (√x)². This simplifies to x, assuming x is nonnegative in accordance with the domain of √x.
For the function f(x) = |x - 3|, what is the value of f(5)?
2
8
-2
0
Substituting x = 5 into f(x) = |x - 3| gives |5 - 3|, which equals 2. This direct evaluation confirms that 2 is the correct answer.
Find the inverse of the function f(x) = (2x + 3)/(x - 1).
f❻¹(x) = (x + 3)/(x - 2)
f❻¹(x) = (x - 3)/(x + 2)
f❻¹(x) = (2x - 3)/(x + 1)
f❻¹(x) = (x + 1)/(2 - x)
To find the inverse, swap x and y in the equation y = (2x + 3)/(x - 1) and solve for y. This process yields f❻¹(x) = (x + 3)/(x - 2), which is the correct inverse function.
Determine the number of real solutions for the equation x❴ - 5x² + 4 = 0.
4
2
0
3
By substituting u = x², the equation transforms to u² - 5u + 4 = 0, which factors as (u - 1)(u - 4) = 0. Solving for u gives u = 1 and u = 4, leading to x = ±1 and x = ±2, a total of four real solutions.
Factor the polynomial f(x) = x³ - 6x² + 11x - 6.
(x - 1)(x - 2)(x - 3)
(x + 1)(x + 2)(x + 3)
(x - 1)²(x - 6)
(x - 2)³
The polynomial factors neatly into (x - 1)(x - 2)(x - 3), which can be verified by either synthetic division or the Rational Root Theorem. This factorization demonstrates that the roots are 1, 2, and 3.
Evaluate the limit: lim(x → ∞) (3x² + 2) / (5x² - 7).
3/5
5/3
0
For rational functions where the highest power in the numerator and denominator is the same, the limit as x approaches infinity is the ratio of the leading coefficients. Thus, the limit is 3/5.
Solve the trigonometric equation 2 sin²(x) - 1 = 0 for 0 ≤ x < 2π.
x = π/4, 3π/4, 5π/4, 7π/4
x = π/4, 7π/4
x = π/2, 3π/2
x = 0, π
Rewriting the equation gives sin²(x) = 1/2, and taking the square root results in sin(x) = ±√2/2. This leads to four solutions in the interval [0, 2π): π/4, 3π/4, 5π/4, and 7π/4.
0
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Study Outcomes

  1. Understand fundamental precalculus concepts and principles.
  2. Analyze functions and their graphical representations.
  3. Apply techniques to solve equations and inequalities.
  4. Evaluate transformations and their effects on function behavior.
  5. Identify areas for improvement to boost mathematical confidence.

Precalculus Unit 3 Test Answers Cheat Sheet

  1. Grasp complex number basics - Complex numbers mix a real part and an imaginary part (a + bi) to extend traditional arithmetic. They help solve equations that otherwise have no real solution and pop up in electronics and signal analysis. Embrace their quirky nature and see math in a whole new light! OpenStax Precalculus - Key Concepts
  2. Operate on complex numbers - Hone your skills adding, subtracting, multiplying, and dividing complex numbers just like binomials. When dividing, multiply by the conjugate to eliminate the imaginary denominator and keep your answers neat. This mastery turns abstract ideas into manageable calculations! OpenStax Precalculus - Key Concepts
  3. Master the quadratic formula - Use x = (-b ± √(b² - 4ac)) / (2a) to crack any quadratic equation, even those with no real roots. The discriminant (b² - 4ac) reveals whether solutions are real, repeated, or complex - your built-in solution detector. Memorize it once, use it forever! OpenStax Precalculus - Key Concepts
  4. Find a quadratic's vertex - In the standard form f(x) = ax² + bx + c, locate the peak or valley at x = -b/(2a). Plug this back into f(x) to get the exact y-coordinate of the vertex. Knowing this makes graphing parabolas a breeze! OpenStax Precalculus - Key Concepts
  5. Analyze end behavior of polynomials - Look at the leading term ax❿: if n is even and a > 0, both ends rise; if a < 0, both fall. If n is odd, the ends go opposite ways depending on the sign of a. This quick check predicts graphs' long-term trends in seconds! OpenStax Precalculus 2e - Key Concepts
  6. Divide polynomials confidently - Choose long division for clarity or synthetic division for speed when dividing by (x - k). Both methods simplify expressions and help pinpoint exact zeros. With practice, these tools become second nature in algebraic adventures! OpenStax Precalculus - Key Concepts
  7. Apply the Remainder Theorem - Divide f(x) by (x - c) and the remainder equals f(c), so you can evaluate polynomials without full division. This theorem saves time and checks your work quickly. It's a true lifesaver on timed exams! OpenStax Precalculus - Key Concepts
  8. Use the Rational Zero Theorem - Predict possible rational roots of a polynomial by listing factors of the constant term over factors of the leading coefficient. Test these candidates with synthetic division or substitution to find real zeros. It's like having a treasure map to your solutions! OpenStax Precalculus - Key Concepts
  9. Solve triangles with the Law of Sines - The ratio sin A/a = sin B/b = sin C/c unlocks unknown sides or angles in non-right triangles. It's perfect for oblique triangles when you have two angles and one side or two sides and a non-included angle. Keep this law handy for geometry puzzles and real-world navigation! OpenStax Precalculus 2e - Trig Key Concepts
  10. Crack triangles with the Law of Cosines - Use c² = a² + b² - 2ab cos C to find missing sides or angles in any triangle, especially when you know two sides and the included angle. It generalizes the Pythagorean theorem and powers up your triangle-solving toolkit. Cosines make the world of triangles a lot less mysterious! OpenStax Precalculus 2e - Trig Key Concepts
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