Unit 1, lesson 7

Reading Graphs and Function Behavior

How to read intercepts, rising and falling intervals, signs and highest and lowest points from a graph, why the scale matters, and when a value read from a picture needs an exact check.

  • Theory part 13
  • Practice part 14
  • Rates and Graph Behavior

Theory, part 13

Practice, part 14

A graph of the temperature over one day answers questions at a glance. When was it warmest? When was it below freezing? When was it warming up? Each question has a precise version in terms of the function, and the precise version is what lets you check what you read instead of trusting your eyes.

Key terms

Intercepts on a graph

Definition The yy-intercept of the graph of ff is the point (0,f(0))(0, f(0)), which exists only when 00 is in the domain. By the vertical line test there is at most one. The xx-intercepts are the points (a,0)(a, 0) with f(a)=0f(a) = 0, and their first coordinates aa are the real zeros of ff. A graph can have no xx-intercepts, one, or many.

In plain words Look where the graph meets the axes. On the yy-axis the input is 00, and on the xx-axis the output is 00.

Example Let gg be the function on [−4,5][-4, 5] whose graph joins the points

(−4,2),(−2,−2),(1,−2),(3,4),(5,0)\begin{gathered} (-4, 2), \quad (-2, -2), \quad (1, -2), \\[0.4em] (3, 4), \quad (5, 0) \end{gathered}

with straight segments. Its yy-intercept is (0,−2)(0, -2), on the flat piece. On the first piece the formula is g(x)=−2x−6g(x) = -2x - 6, which is zero at x=−3x = -3. On the piece from (1,−2)(1, -2) to (3,4)(3, 4) it is g(x)=3x−5g(x) = 3x - 5, which is zero at x=53x = \frac{5}{3}. Together with the endpoint (5,0)(5, 0) this gives the xx-intercepts (−3,0)(-3, 0), (53,0)\left(\frac{5}{3}, 0\right) and (5,0)(5, 0).

Increasing and decreasing intervals

Definition Let II be an interval in the domain of ff. The function ff is increasing on II if f(x1)<f(x2)f(x_1) < f(x_2) whenever x1<x2x_1 < x_2 in II, and decreasing on II if f(x1)>f(x2)f(x_1) > f(x_2) whenever x1<x2x_1 < x_2 in II. Some books call these strictly increasing and strictly decreasing. A few books use increasing for the weaker condition f(x1)≤f(x2)f(x_1) \le f(x_2), which this course calls nondecreasing, as the calculus course does. Increasing and decreasing describe a whole interval, not a single point. An endpoint may be included when the inequality still holds there, so x2x^2 is increasing on [0,∞)[0, \infty). Many books list only open intervals such as (0,∞)(0, \infty), which is also correct. The same condition can be checked on any set of inputs. A function can be decreasing on each of two intervals without being decreasing on their union, so list such intervals separately, and the same goes for increasing.

In plain words Reading from left to right, an increasing graph goes up and a decreasing graph goes down. Report the answer as intervals of inputs on the xx-axis, not as heights.

Example The function gg from the intercepts example is decreasing on [−4,−2][-4, -2] and on [3,5][3, 5], and increasing on [1,3][1, 3]. It is not decreasing on the union [−4,−2]∪[3,5][-4, -2] \cup [3, 5], because −2<3-2 < 3 while g(−2)=−2<g(3)=4g(-2) = -2 < g(3) = 4.

Constant intervals

Definition A function ff is constant on an interval II if f(x1)=f(x2)f(x_1) = f(x_2) for all x1x_1 and x2x_2 in II. Its graph over II is a horizontal segment. On an interval with more than one point, a constant function is neither increasing nor decreasing.

In plain words Where the graph runs flat, the output does not change at all.

Example The function gg is constant on [−2,1][-2, 1], with value −2-2. With the previous example this describes all of gg. It decreases on [−4,−2][-4, -2], stays at −2-2 on [−2,1][-2, 1], increases on [1,3][1, 3] and decreases on [3,5][3, 5].

Positive and negative values

Definition A function ff is positive at xx if f(x)>0f(x) > 0 and negative at xx if f(x)<0f(x) < 0. On the graph, positive values lie above the xx-axis and negative values lie below it. The sets where ff is positive or negative are the solution sets of the inequalities f(x)>0f(x) > 0 and f(x)<0f(x) < 0. If the graph has no breaks over an interval, ff can change sign there only at a zero, so the zeros cut such an interval into pieces of constant sign. A function such as 1x\frac{1}{x} can also change sign at a gap in its domain.

In plain words Positive means above the axis and negative means below. Find the zeros, then test one input in each piece between them.

Example The polynomial p(x)=(x+2)(x−1)(x−4)p(x) = (x + 2)(x - 1)(x - 4) has zeros −2-2, 11 and 44, and its graph has no breaks. One test input in each piece gives the signs.

IntervalTest inputValue
(−∞,−2)(-\infty, -2)−3-3p(−3)=−28<0p(-3) = -28 < 0
(−2,1)(-2, 1)00p(0)=8>0p(0) = 8 > 0
(1,4)(1, 4)22p(2)=−8<0p(2) = -8 < 0
(4,∞)(4, \infty)55p(5)=28>0p(5) = 28 > 0

So pp is positive on (−2,1)∪(4,∞)(-2, 1) \cup (4, \infty) and negative on (−∞,−2)∪(1,4)(-\infty, -2) \cup (1, 4).

Local and global extrema

Definition Let ff have domain DD and let cc be in DD. The value f(c)f(c) is the global (or absolute) maximum of ff if f(c)≥f(x)f(c) \ge f(x) for every xx in DD, and the global minimum if f(c)≤f(x)f(c) \le f(x) for every xx in DD. It is a local (or relative) maximum if there is a number δ>0\delta > 0 such that f(c)≥f(x)f(c) \ge f(x) for every xx in DD with ∣x−c∣<δ\lvert x - c \rvert < \delta, and a local minimum if the same holds with f(c)≤f(x)f(c) \le f(x). Maxima and minima together are called extrema. Here, as in the calculus course, a local extremum may sit at an endpoint of the domain, although some books exclude endpoints. Every global extremum is also a local one. A global maximum value is a single number, but it can be attained at several inputs.

In plain words The global maximum is the greatest height the whole graph reaches. A local maximum only has to be at least as high as the graph nearby, like a hilltop that is lower than the mountain next to it. Keep the value, which is a height, apart from the input where it happens.

Example Take f(x)=x3−6x2+9xf(x) = x^3 - 6x^2 + 9x on the domain [0,4][0, 4].

xxf(x)f(x)
0000
1144
2222
3300
4444

The table suggests a peak at x=1x = 1 and a valley at x=3x = 3, and factoring proves it. For every xx in [0,4][0, 4]

f(x)=x(x−3)2≥0f(x) = x(x - 3)^2 \ge 0

and

4−f(x)=(x−1)2(4−x)≥0.4 - f(x) = (x - 1)^2 (4 - x) \ge 0.

So the global minimum value is 00, attained at x=0x = 0 and x=3x = 3, and the global maximum value is 44, attained at x=1x = 1 and at the endpoint x=4x = 4. All four are also local extrema, and the two interior ones, at x=1x = 1 and x=3x = 3, are where the graph turns around. Calculus gives a systematic way to find such points, in extreme values.

Graph scale

Definition The scale of an axis says how many units of input or output one unit of length along the axis stands for, as marked by its tick labels. A viewing window [xmin⁡,xmax⁡][x_{\min}, x_{\max}] by [ymin⁡,ymax⁡][y_{\min}, y_{\max}] shows only the points of the graph whose coordinates lie in those intervals. Changing the scale or the window changes how steep or flat the graph looks and which features are visible, but it does not change the function. Heights and slopes must be read from the axis labels, not from distances on the screen or page.

In plain words The same function can look steep or flat, wiggly or smooth, depending on the window. Check the numbers on the axes before you trust a shape.

Example The function f(x)=x3−xf(x) = x^3 - x has zeros at −1-1, 00 and 11, a small hump between −1-1 and 00 and a small dip between 00 and 11, with f(−0.5)=0.375f(-0.5) = 0.375 and f(0.5)=−0.375f(0.5) = -0.375. In the window [−1.5,1.5][-1.5, 1.5] by [−1,1][-1, 1] the hump and the dip are plain to see. In the window [−10,10][-10, 10] by [−1000,1000][-1000, 1000] the graph looks just like y=x3y = x^3. The hump there is less than a thousandth of the height of the window, and the three zeros crowd together at the center.

Exact versus estimated values

Definition A value read from a graph is an estimate, accurate only up to the precision of the drawing and its scale. An exact value comes from the formula, by evaluating it or by solving an equation algebraically. A reading such as “the zero is about 1.41.4” is confirmed or corrected by substituting into the formula, and a number aa is an exact zero only if f(a)=0f(a) = 0 exactly.

In plain words Use the graph to see roughly where to look, then use the formula to get the number. A picture cannot tell 1.411.41 from 2\sqrt{2}.

Example The graph of f(x)=x2−2f(x) = x^2 - 2 crosses the positive xx-axis near 1.41.4. But f(1.4)=−0.04f(1.4) = -0.04 and f(1.5)=0.25f(1.5) = 0.25, so 1.41.4 is not a zero, and since the graph has no breaks the crossing lies between 1.41.4 and 1.51.5. Solving x2=2x^2 = 2 gives the exact positive zero 2≈1.41421\sqrt{2} \approx 1.41421.

Check yourself

  1. The graph of ff on [−3,4][-3, 4] joins (−3,−1)(-3, -1), (0,2)(0, 2), (2,2)(2, 2) and (4,−2)(4, -2) with straight segments. Find where ff is increasing, decreasing and constant, its xx-intercepts, and its global extrema.
Show the answer

The function is increasing on [−3,0][-3, 0], constant on [0,2][0, 2] and decreasing on [2,4][2, 4]. The first piece is f(x)=x+2f(x) = x + 2, zero at x=−2x = -2, and the last piece is f(x)=6−2xf(x) = 6 - 2x, zero at x=3x = 3, so the xx-intercepts are (−2,0)(-2, 0) and (3,0)(3, 0). The global maximum value 22 is attained at every xx in [0,2][0, 2]. The global minimum value is −2-2, at the endpoint x=4x = 4, since the other endpoint gives only f(−3)=−1f(-3) = -1.

  1. For h(x)=x2−4xh(x) = x^2 - 4x, find exactly where hh is negative and find its global minimum.
Show the answer

Factoring gives h(x)=x(x−4)h(x) = x(x - 4), which is negative exactly when the two factors have opposite signs, that is on (0,4)(0, 4). Completing the square gives h(x)=(x−2)2−4≥−4h(x) = (x - 2)^2 - 4 \ge -4, with equality only at x=2x = 2. The global minimum value is −4-4, attained at x=2x = 2.

  1. A graph suggests that f(x)=x3+x−1f(x) = x^3 + x - 1 has a zero at x=0.7x = 0.7. Is 0.70.7 an exact zero?
Show the answer

No. The value is f(0.7)=0.343+0.7−1=0.043f(0.7) = 0.343 + 0.7 - 1 = 0.043, which is not 00. Since f(0.68)≈−0.0056<0f(0.68) \approx -0.0056 < 0 and the graph has no breaks, the zero lies between 0.680.68 and 0.70.7. It is about 0.68230.6823.

How this lesson connects