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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.
Study Interpreting Sketching Key Features Of Functions in Algebra with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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For f(x)=−x3, what is the end behavior as x→∞ and x→−∞?
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As x→∞, f(x)→−∞; as x→−∞, f(x)→∞. Odd-degree polynomial with negative leading coefficient.
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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.
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: As x→∞, f(x)→−∞; as x→−∞, f(x)→∞. Odd-degree polynomial with negative leading coefficient.
Answer: (−∞,1)∪(5,∞). Parabola opens upward, so it's positive outside zeros.
Answer: x=3. From (x−3)2, the axis is at x=3.
Answer: As x→∞, f(x)→∞; as x→−∞, f(x)→−∞. Odd-degree polynomial with positive leading coefficient.
Answer: What happens to f(x) as x→∞ and x→−∞. Behavior as x approaches positive and negative infinity.
Answer: Origin symmetry (odd). Since f(−x)=(−x)3=−x3=−f(x), it's odd.
Answer: The point where the graph crosses the y-axis, at x=0. Occurs when the input equals zero.
Answer: f(x)→∞ as x→∞ and as x→−∞. Upward-opening parabola goes to infinity both ways.
Answer: For a<x1<x2<b, f(x1)>f(x2). Larger inputs give smaller outputs throughout the interval.
Answer: A point higher than nearby points (local highest value). A peak on the graph, higher than surrounding points.
Answer: 5. The vertex form shows k is the minimum value.
Answer: (−∞,−2)∪(2,∞). Function is positive outside its zeros at x=±2.
Answer: (1,5). Parabola opens upward, so it's negative between zeros.
Answer: 7. The y-intercept is the output value when input is zero.
Answer: It becomes f(0)−3. Subtracting 3 from f(0) gives the new intercept.
Answer: All real numbers, (−∞,∞). Linear functions with positive slope increase everywhere.
Answer: Shift left 2 units. Adding inside parentheses shifts left, not right.
Answer: −6. Substitute x=0 into f(x)=3(0)−6=−6.
Answer: f(x)→−∞ as x→∞ and as x→−∞. Downward-opening parabola goes to negative infinity both ways.
Answer: It is the output when x=0, the ordered pair (0,f(0)). Find the row where input is zero.
Answer: f(x). Since period is 6, f(x+6)=f(x) by definition.
Answer: x=−2 and x=2. Set f(x)=0: x2−4=0, so x=±2.
Answer: −4. Substitute x=0: f(0)=02−4=−4.
Answer: A point lower than nearby points (local lowest value). A valley on the graph, lower than surrounding points.
Answer: Vertical shift up k units (down if k<0). Adds constant to all output values.
Answer: x=h. Vertical line through the vertex of a parabola.
Answer: As x increases, f(x) decreases on that interval. The function values fall as you move right.
Answer: f(x). Since 12=3×4, we get f(x+12)=f(x).
Answer: f(x)<0 on that interval (graph is below the x-axis). All function values are less than zero.
Answer: Horizontal shift right h units (left if h<0). Replaces x with x−h in the function.
Answer: 2. The vertex form shows h is the x-coordinate.
Answer: As x increases, f(x) increases on that interval. The function values rise as you move right.
Answer: 4. Negative coefficient creates downward parabola with maximum k.
Answer: For a<x1<x2<b, f(x1)<f(x2). Larger inputs give larger outputs throughout the interval.
Answer: Symmetric about the y-axis; f(−x)=f(x). Folding across y-axis gives identical graph.
Answer: x=−4. From (x+4)2, the axis is at x=−4.
Answer: Symmetric about the origin; f(−x)=−f(x). Rotating 180° about origin gives identical graph.
Answer: −3. The x-intercept is the input value when output is zero.
Answer: 7. The y-intercept is the output value when input is zero.
Answer: −1. The vertex occurs at x=h from the form.
Answer: 2. Set f(x)=0: 3x−6=0, so x=2.
Answer: f(x)>0 on that interval (graph is above the x-axis). All function values are greater than zero.
Answer: Intercepts, extrema, increasing/decreasing, sign, end behavior, period. Essential elements for complete function analysis.
Answer: The smallest P>0 such that f(x+P)=f(x). Adding the period returns the same output.
Answer: All real numbers, (−∞,∞). Linear functions with negative slope decrease everywhere.
Answer: y-axis symmetry (even). Since f(−x)=(−x)2−1=x2−1=f(x), it's even.
Answer: A repeating pattern with some positive period P. Function values repeat at regular intervals.
Answer: It occurs where the output is 0, at a point (x,0). Find the row where output is zero.
Answer: A point where the graph crosses the x-axis, where y=0. Occurs when the output equals zero.
Answer: x=1 and x=5. Set each factor to zero: x−1=0 or x−5=0.
Answer: x=3. From (x−3)2, the axis is at x=3.
Answer: −3. The x-intercept is the input value when output is zero.
Answer: f(x). Since 12=3×4, we get f(x+12)=f(x).
Answer: (−2,2). Function is negative between its zeros at x=±2.