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What does it mean for a relation to be a function, and how do you read the key features of its graph?

Understand the definition of a function and function notation; evaluate functions; identify domain and range; and interpret the key features of a graph (intercepts, intervals of increase and decrease, relative maxima and minima, and average rate of change) in context.

A NY Regents Algebra I answer on functions: the definition and the vertical-line test, function notation and evaluation, domain and range, and reading key features of a graph such as intercepts, increasing intervals, and average rate of change.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. What makes a relation a function
  3. Function notation and evaluation
  4. Domain and range in context
  5. Key features of a graph
  6. Connecting features to context
  7. Try this

What this topic is asking

The Regents Algebra I exam (the Interpreting Functions, F-IF, cluster) wants you to know what a function is, use function notation to evaluate, identify domain and range, and read the key features of a graph: intercepts, where it increases or decreases, relative maxima and minima, and the average rate of change over an interval. These ideas appear throughout Part I and in many constructed-response questions.

What makes a relation a function

A function pairs each input with exactly one output. The set of inputs is the domain; the set of outputs is the range. On a graph, the vertical-line test decides it: if any vertical line crosses the graph more than once, some input has two outputs, so it is not a function. A circle fails the test; a parabola opening up passes it.

In a table, a relation is a function unless an input (xx-value) repeats with different outputs. The input 5→25 \to 2 and 5→75 \to 7 would disqualify it.

Function notation and evaluation

The notation f(x)f(x) reads "ff of xx" and names the output of the function ff at input xx. It is not multiplication. To evaluate, substitute the input everywhere xx appears and simplify. For g(x)=x2−4xg(x) = x^2 - 4x, g(3)=9−12=−3g(3) = 9 - 12 = -3. You may also solve f(x)=kf(x) = k for the input that produces a given output, reading it as "for what xx is the output kk".

Domain and range in context

The domain is every input the function allows, and the range is every output it produces. For a pure expression like f(x)=xf(x) = \sqrt{x} the domain excludes negatives. In a context, the domain is restricted to values that make sense: a function giving the number of bacteria after tt hours has domain t≥0t \geq 0, because negative time is meaningless. The Regents often asks for an "appropriate domain", which means the realistic restriction, not all real numbers.

Key features of a graph

Connecting features to context

In a real situation each feature has meaning. The yy-intercept is the starting value, a zero is when the quantity reaches zero (a ball landing, a balance emptying), a relative maximum is a peak (greatest height, maximum profit), and the average rate of change is the overall pace (average speed, average growth). A clarifying point is that the average rate of change between two points is generally not the same as the rate at a single instant; it is the slope of the straight line joining the endpoints, which is exactly what the Regents asks for when it says "average rate of change over the interval".

Try this

Q1. If f(x)=5−2xf(x) = 5 - 2x, find f(4)f(4). [1 credit]

  • Cue. f(4)=5−8=−3f(4) = 5 - 8 = -3.

Q2. A graph of distance versus time passes through (2,10)(2, 10) and (6,30)(6, 30). Find the average rate of change. [2 credits]

  • Cue. 30−106−2=204=5\frac{30 - 10}{6 - 2} = \frac{20}{4} = 5 units per unit of time.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart I (multiple choice). If f(x)=2x2−3x+1f(x) = 2x^2 - 3x + 1, what is the value of f(−2)f(-2)? (1) 33 (2) 1515 (3) −1-1 (4) 1111
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The correct answer is (2).

Substitute x=−2x = -2: f(−2)=2(−2)2−3(−2)+1=2(4)+6+1=8+6+1=15f(-2) = 2(-2)^2 - 3(-2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = 15. The most common error is mishandling (−2)2(-2)^2, treating it as −4-4 instead of 44, which gives −1-1 (choice 3). The square of a negative is positive.

Regents (style)2 marksPart II (constructed response). The graph of y=f(x)y = f(x) passes through (1,4)(1, 4) and (5,12)(5, 12). Calculate the average rate of change of ff over the interval 1≤x≤51 \leq x \leq 5, and state its units if xx is in seconds and yy is in meters.
Show worked answer →

A 2-credit question: 1 credit for the value, 1 for the units.

Average rate of change =f(5)−f(1)5−1=12−44=84=2= \frac{f(5) - f(1)}{5 - 1} = \frac{12 - 4}{4} = \frac{8}{4} = 2. With xx in seconds and yy in meters, the units are meters per second. A bare "2" with no units, or computing the change in xx over the change in yy by accident, costs a credit.

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