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This deck focuses on Interpreting Sketching Key Features Of Functions, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.
Study Interpreting Sketching Key Features Of Functions in Algebra 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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Which interval is f(x)=x2−4 positive on?
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(−∞,−2)∪(2,∞). Parabola is positive outside the roots.
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This deck focuses on Interpreting Sketching Key Features Of Functions, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: (−∞,−2)∪(2,∞). Parabola is positive outside the roots.
Answer: As x increases, f(x) increases on that interval. The graph slopes upward from left to right.
Answer: Average rate of change is 5−27−1=2. Formula: change in inputchange in output.
Answer: Decreasing on (−1,∞). Right side of downward-opening parabola.
Answer: The vertex is (h,k). The turning point of the parabola.
Answer: (2,5). Parabola is negative between its roots.
Answer: As x→∞, f(x)→−∞; as x→−∞, f(x)→∞. Odd degree with negative leading coefficient.
Answer: It satisfies f(−x)=−f(x) and has origin symmetry. Point reflection through the origin.
Answer: Increasing on (3,∞). Right side of upward-opening parabola.
Answer: How f(x) behaves as x→∞ and as x→−∞. Describes the function's behavior at extreme x-values.
Answer: Maximum value is 7. Vertex gives the maximum for downward parabola.
Answer: The relative minimum is at (3,5). Parabola opens upward with vertex at (3,5).
Answer: As x increases, f(x) decreases on that interval. The graph slopes downward from left to right.
Answer: For all x in (a,b), f(x)>0. All function values are above the x-axis.
Answer: A point where f(x) is greatest compared to nearby x-values. A local peak on the graph.
Answer: The y-intercept is (0,−6). Evaluate f(0)=2(0)−6=−6.
Answer: The x-intercept is (3,0). Set 2x−6=0 and solve for x.
Answer: The x-intercepts are (1,0) and (−4,0). Set each factor equal to zero.
Answer: The midline is y=2. Vertical shift of the sine function.
Answer: Horizontal asymptote: y=0. Rational function approaches zero as x grows.
Answer: The period is b2π. Coefficient b compresses the period by factor b.
Answer: There exists T>0 such that f(x+T)=f(x) for all x. The function repeats its pattern every T units.
Answer: (−∞,2)∪(5,∞). Parabola is positive outside its roots.
Answer: (−2,2). Parabola is negative between the roots.
Answer: The point where x=0, so the value is (0,f(0)). Where the graph crosses the y-axis.
Answer: The y-intercept is (0,−4). Evaluate f(0)=(0−1)(0+4)=−4.
Answer: The period is 2π. Standard period for cosine function.
Answer: Increasing on (−∞,−1). Left side of downward-opening parabola.
Answer: The period is 2π. Period formula: 42π=2π.
Answer: A point where f(x) is least compared to nearby x-values. A local valley on the graph.
Answer: Vertical asymptote: x=3. Denominator cannot equal zero.
Answer: Odd; it has origin symmetry. Standard reciprocal function has origin symmetry.
Answer: The period is 6π. Period formula: 1/32π=6π.
Answer: The graph is above the x-axis at x=a. Positive function values mean points above the axis.
Answer: As x→±∞, f(x)→0. Rational function approaches horizontal asymptote.
Answer: The amplitude is 3. Coefficient of sine function.
Answer: The axis of symmetry is x=h. Vertical line through the vertex of the parabola.
Answer: The vertex is (3,1). From vertex form (x−3)2+1.
Answer: The x-intercepts (zeros) of the function. Found by setting the function equal to zero.
Answer: The y-intercept is (0,1). Evaluate f(0)=0+1=1.
Answer: Domain: x≥4. Expression under square root must be non-negative.
Answer: As x→±∞, f(x)→∞. Even degree with positive leading coefficient.
Answer: A point where y=0 on the graph, so f(x)=0. Where the graph crosses the x-axis.
Answer: The graph is below the x-axis at x=a. Negative function values mean points below the axis.
Answer: It is the slope of the secant line: b−af(b)−f(a). Geometric interpretation of average rate of change.
Answer: Decreasing on (−∞,3). Left side of upward-opening parabola.
Answer: It satisfies f(−x)=f(x) and is symmetric about the y-axis. Mirror image across the y-axis.
Answer: The period is 2π. Standard period for sine function.
Answer: The y-intercept (0,f(0)). Found by evaluating the function at zero.
Answer: When input is a, the output is f(a)=b. Input-output relationship in tabular form.
Answer: The period is b2π. Coefficient b compresses the period by factor b.
Answer: If x1<x2 in (a,b), then f(x1)<f(x2). Definition using ordered pairs in the interval.
Answer: The relative maximum is at (−1,7). Parabola opens downward with vertex at (−1,7).
Answer: Neither. Neither even nor odd function.
Answer: The axis of symmetry is x=3. Complete the square: (x−3)2+1.