2.1 Functions, domains, graphs, and rates of change - Courses of Computing4All
College Algebra for High School Students 2.1 Functions, domains, graphs, and rates of change
College Algebra for High School Students
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Learning goals: Recognize functions; evaluate and interpret notation; determine domain and range; compute average rates of change.

One input, exactly one output

A function assigns exactly one output to each input in its domain. Several inputs may share an output. For example, \(f(x)=x^2\) sends both three and negative three to nine; that is still a function. A relation containing \((2,5)\) and \((2,7)\) is not a function of the first coordinate because the input two has competing outputs. On a graph, this is the vertical line test: no vertical line may intersect the graph more than once.

See the vertical line test

Vertical line test: x equals one meets the parabola once and the circle twice.−3−2−1012302468xy(1, 1)Function: y = x²x = 1−3−2−10123−3−2−10123xyNot a function of x: x² + y² = 4x = 1One output for each inputOne input gives two outputs

Reading the graph. The dashed line fixes the input at \(x=1\). On \(y=x^2\), it meets only \((1,1)\). Every other vertical line also meets this parabola exactly once. On the circle, the same input gives \(y=\sqrt3\) and \(y=-\sqrt3\), so the circle is not a function of x. Two different inputs sharing an output is allowed; one input having two outputs is not. The parabola continues beyond the displayed window.

Pause and explain: The points \((-2,4)\) and \((2,4)\) are both on the parabola. Do they make it fail the test?

Check your reasoning

No. They have different inputs. The test compares outputs for the same input, so these two points do not conflict.

Function notation is an instruction

The notation \(f(3)\) means the output of \(f\) at input three; it does not mean multiplication. For \(f(x)=2x^2-3x+1\), \[f(-2)=2(-2)^2-3(-2)+1=8+6+1=15.\] To evaluate \(f(a+h)\), replace every occurrence of \(x\): \[f(a+h)=2(a+h)^2-3(a+h)+1.\] Keep parentheses around substituted expressions before expanding. Solving \(f(x)=15\) reverses the question: find all inputs that produce fifteen.

Domain and range

The domain is the set of allowed inputs; the range is the set of outputs actually attained. For a real-valued formula, begin by excluding zero denominators and negative quantities under even roots. The function \(g(x)=\frac{\sqrt{x+2}}{x-3}\) requires \(x\ge-2\) and \(x\ne3\), so its domain is \([-2,3)\cup(3,\infty)\). These restrictions must be combined, not considered separately.

For \(y=(x-1)^2+2\), all real inputs are allowed. A square is nonnegative, so the least output is two, attained at \(x=1\), and the range is \([2,\infty)\). Domain usually comes from allowable inputs; range often requires examining the graph or the formula’s structure. Context may impose extra restrictions such as time being nonnegative.

Read domain and range in different directions

An upward parabola has its lowest point at one, two; the dashed horizontal line marks the minimum output two.−3−2−1012345024681012xyMinimum: (1, 2)(0, 3)y = (x − 1)² + 2Domain: all real xRange: y ≥ 2, including 2

Reading the graph. Read inputs horizontally and outputs vertically. The graph continues for every real input, so its domain is \(( -\infty,\infty)\). Its vertex \((1,2)\) is included, and every output above two occurs, so its range is \([2,\infty)\). The dashed line marks the lowest output; it is a guide, not another part of the function. The edges of the drawing are only a viewing window, not endpoints of the parabola.

Pause and explain: Does this function ever output zero? Why is two included in its range?

Check your reasoning

It never outputs zero because \((x-1)^2+2\ge2\). It outputs two at \(x=1\), so the range includes its lower endpoint.

What a graph tells us

An x-intercept satisfies \(f(x)=0\); the y-intercept is \(f(0)\), if zero belongs to the domain. A function is increasing on an interval when larger inputs give larger outputs throughout that interval. It is decreasing when larger inputs give smaller outputs. A high point is not automatically the absolute maximum: compare all allowed inputs, including endpoints when present. Open endpoints are excluded, and a function can approach a bound without attaining it.

Identify intercepts, direction, and endpoints

Left: the parabola x squared minus four crosses the x axis at minus two and two and the y axis at minus four. Right: a rising line segment has an open left endpoint and a filled right endpoint.−3−2−10123−4−20246xy(−2, 0)(2, 0)(0, −4)Intercepts: f(x) = x² − 4−2−101234−2−101234xy(−1, −1)(3, 3)Endpoints: h(x) = x, −1 < x ≤ 3Decreasing, then increasingOpen = excluded; filled = included

Reading the graph. On the left, \((-2,0)\) and \((2,0)\) are x-intercepts, while \((0,-4)\) is the y-intercept and absolute minimum. As you move left to right, the function decreases on \((-\infty,0)\) and increases on \((0,\infty)\). It has no absolute maximum. On the right, the restricted function has domain \((-1,3]\) and range \((-1,3]\). Its maximum is three. It has no minimum: outputs approach negative one, but the open point excludes negative one itself.

Pause and explain: For the right-hand graph, is the lowest output \(-1\)? Would the answer change if that endpoint were filled?

Check your reasoning

No lowest output is attained in the graph shown. For any output above \(-1\), there is an allowed output closer to \(-1\). Filling the endpoint would include \((-1,-1)\), so \(-1\) would become the minimum.

Average rate of change

Between distinct inputs \(a\) and \(b\), the average rate is \[\frac{f(b)-f(a)}{b-a}.\] It is the slope of the secant line joining the two graph points. For \(f(x)=x^2\) from one to four, the rate is \((16-1)/(4-1)=5\). This is not the output at the midpoint; it compares changes in outputs with changes in inputs. If inputs are seconds and outputs meters, the units are meters per second. Linear functions have a constant rate on every interval, while nonlinear functions generally do not.

See why average rate is a secant slope

The secant line joins one, one and four, sixteen on y equals x squared. A dashed right triangle shows a run of three and a rise of fifteen.01234048121620xyA = (1, 1)B = (4, 16)Run = 4 − 1 = 3Rise = 15Curve: f(x) = x²Secant line: y = 5x − 4Average rate = rise ÷ run = 15 ÷ 3 = 5

Reading the graph. The blue curve gives the actual outputs of \(f(x)=x^2\). The magenta straight line passes through \(A=(1,1)\) and \(B=(4,16)\). The dashed triangle measures an input change of three and an output change of fifteen, so the secant slope is \(15/3=5\). The triangle is a measuring aid, not another function. The curve is not the same as the secant between A and B: its rate varies, while five summarizes the whole interval.

Pause and explain: From \(x=0\) to \(x=1\), is the average rate still five? Use two graph points to explain.

Check your reasoning

No. The points are \((0,0)\) and \((1,1)\), so the average rate is \((1-0)/(1-0)=1\). Different intervals can have different secant slopes on a nonlinear graph.

Practice: attempt before reading the solutions

Show your reasoning, state restrictions, and check your result. For explanation questions, use complete mathematical sentences.

  1. Is \(\{(1,4),(2,4),(3,9)\}\) a function?
  2. If \(f(x)=3x^2-2\), find \(f(-2)\).
  3. Find the real domain of \(1/(x^2-9)\).
  4. Find the real domain of \(\sqrt{7-2x}\).
  5. Find the range of \((x+3)^2-5\) on all real inputs.
  6. Find the average rate of \(f(x)=x^2+1\) from two to five.
  7. For \(f(x)=x^2-4\), identify the intercepts.
  8. Explain the difference between \(f(3)=8\) and \(f(x)=8\).

Worked solutions and reasoning

  1. Yes. Each input appears with exactly one output. Repeating the output four for different inputs is allowed.
  2. \(f(-2)=3(-2)^2-2=12-2=10\). Square the entire negative input before multiplying.
  3. The denominator factors as \((x-3)(x+3)\), so exclude \(-3\) and \(3\). All other real inputs are permitted.
  4. Require \(7-2x\ge0\). Thus \(x\le7/2\), giving \((-\infty,7/2]\).
  5. The square is at least zero, and it equals zero at \(x=-3\). The range is \([-5,\infty)\).
  6. \(f(5)=26\) and \(f(2)=5\). The rate is \((26-5)/(5-2)=21/3=7\).
  7. Set the output to zero: \(x^2-4=0\), so the x-intercepts are \((-2,0),(2,0)\). At \(x=0\), the y-intercept is \((0,-4)\).
  8. The first reports an output for a specified input. The second is an equation whose solutions are all inputs that produce eight; there may be several.

Reflection

Choose one problem you initially found difficult. Explain the key idea, identify a possible mistake, and write a new problem that uses the same idea. Solve your new problem and check it independently.

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