Functions and Their Graphs — Free Pre-Calculus Review Games.
This unit covers domain and range, transformations, piecewise functions and composition of functions — essential concepts for Pre-Calculus. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers domain and range, transformations, piecewise functions and composition of functions — essential concepts for Pre-Calculus. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 Domain And Range
Domain is the set of all valid input values (x) for a function, and range is the set of all valid output values (y). Students must identify restrictions caused by square roots (radicand ≥ 0), fractions (denominator ≠ 0), and logarithms (argument > 0). Answers are expressed in interval notation or set-builder notation.
Key Points
- For fractions: set denominator ≠ 0 and solve for excluded x-values
- For even roots: set the radicand ≥ 0 and solve the inequality
- For range: analyze the graph or solve the function for x in terms of y, then find valid y-values
- Interval notation: use parentheses for strict inequalities, brackets for ≤ or ≥
Find the domain of f(x) = √(2x − 6) / (x − 5)
The square root requires 2x − 6 ≥ 0, giving x ≥ 3. The denominator requires x − 5 ≠ 0, giving x ≠ 5. Combining both restrictions, the domain is [3, 5) ∪ (5, ∞).
2 Transformations
Transformations shift, reflect, stretch, or compress a parent function's graph in predictable ways based on changes to the equation. Students must correctly identify whether a change affects the graph horizontally (inside the function, opposite direction) or vertically (outside the function, same direction). Knowing the parent functions (quadratic, absolute value, square root, cubic) is required.
Key Points
- f(x) + k shifts UP k; f(x) − k shifts DOWN k (vertical, same direction)
- f(x − h) shifts RIGHT h; f(x + h) shifts LEFT h (horizontal, opposite direction)
- −f(x) reflects over the x-axis; f(−x) reflects over the y-axis
- a·f(x) with |a| > 1 is a vertical stretch; 0 < |a| < 1 is a vertical compression
Describe the transformations applied to f(x) = x² to obtain g(x) = −2(x + 3)² − 1
The (x + 3) shifts the graph left 3 units. The factor of −2 reflects it over the x-axis and vertically stretches it by a factor of 2. The − 1 at the end shifts the graph down 1 unit. The vertex moves from (0, 0) to (−3, −1).
3 Piecewise Functions
A piecewise function uses different expressions for different intervals of the domain, and students must apply the correct piece based on the input value. On exams, students are asked to evaluate piecewise functions at specific values, graph them, or determine continuity at boundary points. Checking whether endpoints are included (closed vs. open dots on a graph) is frequently tested.
Key Points
- To evaluate f(a), first identify which interval a belongs to, then substitute into that expression only
- At a boundary x = c, check if left-hand and right-hand pieces give the same output to determine continuity
- Graph each piece only over its stated interval; use open circles for strict inequalities, closed for ≤ or ≥
- A piecewise function can still pass the vertical line test and be a valid function
Given f(x) = { x² + 1, x < 2 ; 3x − 1, x ≥ 2 }, find f(2) and f(−1), then determine if f is continuous at x = 2
For f(2): since 2 ≥ 2, use 3(2) − 1 = 5. For f(−1): since −1 < 2, use (−1)² + 1 = 2. To check continuity at x = 2, the left piece gives 2² + 1 = 5 and the right piece gives 3(2) − 1 = 5; both equal 5, so f is continuous at x = 2.
4 Composition Of Functions
Composition of functions means substituting one entire function into another, written as (f ∘ g)(x) = f(g(x)), where g is applied first and f is applied second. Students must correctly substitute the full expression for g(x) into f and then simplify. The domain of the composition is restricted to x-values in the domain of g whose outputs are also in the domain of f.
Key Points
- (f ∘ g)(x) = f(g(x)): apply g first, then f — order matters and is not commutative in general
- To find (f ∘ g)(a) numerically: compute g(a) first, then plug that result into f
- For the domain of f ∘ g: start with the domain of g, then also exclude values where g(x) falls outside the domain of f
- Decomposing a composite function (finding f and g such that h(x) = f(g(x))) is a common exam task
Let f(x) = 2x + 1 and g(x) = x² − 3. Find (f ∘ g)(x) and evaluate (f ∘ g)(4).
First, substitute g(x) into f: (f ∘ g)(x) = f(g(x)) = 2(x² − 3) + 1 = 2x² − 6 + 1 = 2x² − 5. To evaluate at x = 4: (f ∘ g)(4) = 2(4)² − 5 = 2(16) − 5 = 32 − 5 = 27.
Questions, answered.
What is Functions and Their Graphs?
Functions and Their Graphs is Unit 1 of Pre-Calculus, covering domain and range, transformations, piecewise functions and composition of functions.
How to study for Pre-Calculus Unit 1?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 28+ review questions across 5 different game modes.