Quadratic functions in context
Someone hands you a quadratic that describes something real. What does each part of its graph say about that thing?
Where you are 3 in Quadratic functions
When you reach for it
Someone gives you a quadratic that describes something real: a leaping dolphin, a firework, the area inside a loop of rope. It may arrive as a rule, a table or a graph. You already know how to find its features: the intercepts, the vertex and the axis of symmetry, and the vertex as a highest or lowest point. In a situation each feature has a meaning, with units. Reach for this skill when the question is “what does this number say about the situation?” rather than “what is this number?”.
Nothing here is solved by algebra. Every intercept is read from a graph, a table, or a rule given in factored or vertex form. Exact solving comes later, with factoring and the quadratic formula.
The procedure
One made-up model runs through this section. Let t be the time in seconds since a dolphin began its run upward, and h its height in meters relative to the water surface: positive above, negative below. The model is
The two forms agree, since (t - 0.2)(t - 1.4) = t^2 - 1.6t + 0.28, and -5 times that is -5t^2 + 8t - 1.4.
The sign of a: the shape of the story
Here a = -5 < 0, so the parabola opens downward. In words: the height rises, reaches one top, then falls. With a > 0 the quantity would fall to one low point, then rise. The size of a says how sharply the arch bends, not where its top is.
The vertical intercept: the starting value
The vertical intercept is the output when the input is 0: the graph’s y-intercept, here on the h-axis. Here h(0) = -1.4: when the clock starts, the dolphin is 1.4 meters below the surface. When the input is time, this is the starting value, in the output’s units.
The horizontal intercepts: when the quantity is zero
The horizontal intercepts are the graph’s x-intercepts, here on the t-axis. The height is 0 at the surface, because height is measured from the surface. A product is 0 exactly when a factor is, so the factored form shows the zeros: t = 0.2 and t = 1.4. The dolphin leaves the water at 0.2 seconds and goes back in at 1.4 seconds. Both zeros mean something here; a zero at a negative time or a negative length never does.
The axis and the vertex: the peak, and when
The axis of symmetry sits midway between the zeros:
The vertex is on it, at height h(0.8) = -5(0.6)(-0.6) = 1.8. So the vertex (0.8,\ 1.8) says: the top of the leap is 1.8 meters above the water, 0.8 seconds after the start.
The axis says heights are equal at equal times either side of 0.8 seconds: the dolphin is at 1 meter at t = 0.4 and at t = 1.2. Within the leap every height from 0 up to (not including) 1.8 meters is passed twice; the peak is reached once.
The reasonable domain and range
The formula accepts any real t; the situation does not. The model describes the leap from t = 0 until the dolphin is back in the water at t = 1.4. After that the water slows it, while the rule would have it sink faster forever. So a reasonable domain is
The range is every height the model gives on that domain. The highest is 1.8, at the vertex; the lowest is at the start, h(0) = -1.4. So the range is -1.4 \le h \le 1.8, not the h \le 1.8 of the bare formula.
| Feature | Value | What it says about the dolphin |
|---|---|---|
| sign of a | a = -5 < 0 | rises, then falls |
| vertical intercept | h(0) = -1.4 | starts 1.4 m below the surface |
| horizontal intercepts | t = 0.2, t = 1.4 | leaves the water at 0.2 s, back in at 1.4 s |
| axis | t = 0.8 | equal heights at equal times either side of 0.8 s |
| vertex | (0.8,\ 1.8) | top of the leap, 1.8 m, at 0.8 s |
| domain | 0 \le t \le 1.4 | the leap itself |
| range | -1.4 \le h \le 1.8 | every height it passes through |
Worked examples
A plain rule, with nothing to trim it
Read the features of g(x) = 2(x + 2)(x - 4), which describes no situation, and give its domain and range.
SolutionThe leading terms multiply to 2 \cdot x \cdot x = 2x^2, so a = 2 > 0: the graph opens upward, falling to a lowest point and then rising.
The vertical intercept is g(0) = 2(2)(-4) = -16. A product is zero when a factor is, so the horizontal intercepts are x = -2 and x = 4.
The axis is midway, \displaystyle x = \frac{-2 + 4}{2} = 1, and the vertex is g(1) = 2(3)(-3) = -18, the point (1, -18), a lowest point. Check by symmetry: x = 0 and x = 2 are both 1 from the axis, and g(2) = 2(4)(-2) = -16 = g(0).
With no situation, nothing trims the inputs: the domain is all real numbers, and the range is y \ge -18.
A firework shell from a table
Jory’s display team records the height h, in meters above the ground, of a firework shell t seconds after launch (made-up data).
| t (s) | h (m) | t (s) | h (m) | |
|---|---|---|---|---|
| 0 | 1.5 | 4 | 61.5 | |
| 1 | 31.5 | 5 | 51.5 | |
| 2 | 51.5 | 6 | 31.5 | |
| 3 | 61.5 | 7 | 1.5 |
(a) What is the launch height? (b) When is the shell highest, and roughly how high? (c) The team’s rule is h(t) = -5t^2 + 35t + 1.5. Find the peak exactly. (d) The shell bursts at its peak. Give a reasonable domain and range. (e) The rule is zero at about t = 7.04. Find its other zero, and say what each zero means.
Solution(a) The column t = 0 is the starting value: launched 1.5 meters up.
(b) The heights match in pairs: 61.5 at t = 3 and 4, 51.5 at 2 and 5, and so on. Every pair is centered on t = 3.5, so the axis is t = 3.5: the shell is highest at 3.5 seconds. That time is not in the table, so the peak is a little above 61.5 meters. The changes 30, 20, 10, 0, -10, -20, -30 tell the story of a < 0: the shell gains less height each second, then loses height.
(c) On the axis,
The peak is 62.75 meters, at 3.5 seconds.
(d) The model describes the shell from launch to burst, so 0 \le t \le 3.5, and the height climbs from 1.5 to 62.75: the range is 1.5 \le h \le 62.75.
(e) The zeros sit symmetrically about the axis. 7.04 is 3.54 right of 3.5, so the other zero is 3.5 - 3.54 = -0.04, a time before launch: meaningless. The zero at 7.04 is when an unburst shell would land (the table agrees: h(7) = 1.5 is just above the ground). This shell bursts at 3.5 seconds, so 7.04 lies outside its domain.
Estimating from a graph
Use Figure 1 to estimate when the dolphin is 1 meter above the surface, and for how long it stays above that height. Check with the rule h(t) = -5t^2 + 8t - 1.4.
SolutionThe level line at height 1 meets the arch at about t = 0.4 and t = 1.2, both 0.4 from the axis t = 0.8, as symmetry demands. Substituting turns the estimate into a fact:
The dolphin is above 1 meter for 1.2 - 0.4 = 0.8 seconds. Graph readings are seldom this exact.
Writing the rule: a roped rectangle
Keturah has 36 meters of rope to mark out a rectangle on the beach for volleyball, using all of it for the four sides. Write the area A, in square meters, as a function of the width w, in meters, and say what each feature means.
SolutionTwo widths and two lengths use 36 meters, so one of each uses 18: the length is 18 - w, and
| w (m) | A (m²) | w (m) | A (m²) | |
|---|---|---|---|---|
| 0 | 0 | 12 | 72 | |
| 3 | 45 | 15 | 45 | |
| 6 | 72 | 18 | 0 | |
| 9 | 81 |
Sign of a. a = -1 < 0: as the width grows, the area rises, then falls.
Zeros. A = 0 at w = 0 (no width) and w = 18 (no length). Both are rope lying in a doubled line: the edges of the situation, not rectangles.
Axis. w = 9. Widths 6 and 12 both give 72 square meters, since a 6 by 12 rectangle is a 12 by 6 one turned round. The graph’s symmetry is the swap of width and length.
Vertex. (9,\ 81): a 9 by 9 square encloses the most, 81 square meters. Here it is read off, not proved best.
Domain and range. A real rectangle needs 0 < w < 18, and then 0 < A \le 81.
A second shell, given in vertex form
Hallam’s shell follows h(t) = -5(t - 4)^2 + 80, in meters above the ground, t seconds after launch. Read its peak, its launch height, and when it would land if it did not burst. Compare it with Example 2.
Solution(t - 4)^2 is never negative, so h(t) \le 80, with equality only at t = 4: the vertex (4,\ 80) is a peak of 80 meters at 4 seconds.
The starting value is h(0) = -5(16) + 80 = 0: launched from the ground. On the axis t = 4, the time t = 0 is 4 to the left, so t = 8 has the same height: h(8) = -5(16) + 80 = 0. An unburst shell lands after 8 seconds; both zeros mean something, launch and landing.
Expanding, h(t) = -5t^2 + 40t, so a = -5, as in Example 2. Both arches bend equally sharply, one graph the other slid to a new place. Only the vertex differs: Hallam’s shell starts lower (0 against 1.5 meters) and peaks later and higher (80 meters at 4 seconds, against 62.75 at 3.5).
Boundaries
- The rule outlives the situation. Outside the reasonable domain the formula still gives numbers that describe nothing: a dolphin sinking forever, a negative time.
- The vertex is the situation’s peak only if its input is in the domain. Had Jory’s shell burst at 2 seconds, the highest point would be h(2) = 51.5, at the end of the domain.
- A zero means “none of the quantity” only if 0 is meaningful. Measured from the sea floor, the same leap would have different zeros.
- Readings between table values or off a graph are estimates. With a rule, substitute to check.
- Rising then falling does not make a curve a parabola. A table with equal input steps fits a quadratic exactly when its second differences (the changes in the changes) are constant and not zero.
Where this goes next
Finding the best value by algebra from the words of a problem is Maximum, minimum and projectile problems. Exact intercepts of an unfactored rule come with Solving quadratics by factoring and Using the quadratic formula. Fitting the rule to data yourself is Fitting a quadratic to data. The rise-then-fall reading extends to any function in Increasing, decreasing, constant, positive and negative intervals and Turning points, maximums and minimums.
Exercises
Reading a rule
- For f(x) = -(x - 2)(x - 6), find the direction it opens, the horizontal intercepts, the vertical intercept, the axis of symmetry, the vertex and the range.
- For p(x) = (x - 3)^2 - 4, give the vertex and the vertical intercept. One horizontal intercept is x = 1; use symmetry to find the other.
- A second dolphin’s height, in meters relative to the surface, is h(t) = -5(t - 0.3)(t - 1.1), t seconds after its run starts. How deep is it at the start? How long is it out of the water? When is it highest, and how high? Give a reasonable domain and range.
- For the dolphin model h(t) = -5t^2 + 8t - 1.4, say in a sentence with units what h(0.5) = 1.35 means, and check it.
Reading a table
- Hallam records a small shell (made-up data). At t = 0, 0.5, 1, 1.5, 2, 2.5, 3 seconds its heights are 1, 7.25, 11, 12.25, 11, 7.25, 1 meters. Give the launch height, the time and height of the peak, and when it is back at its launch height.
- Hallam’s rule for that shell is h(t) = -5t^2 + 15t + 1, zero at about t = 3.07. Find the other zero by symmetry, say which zero means something, and give a reasonable domain and range for a shell that does not burst.
- A quadratic function f, describing no situation, has f(1) = 7, f(2) = 2, f(3) = -1, f(4) = -2, f(5) = -1. Find the axis and the vertex, say whether the vertex is a maximum or a minimum, predict f(6) and f(7), and say between which whole numbers the horizontal intercepts lie.
Reading a graph
- From Figure 1, during which times is the dolphin above the surface, and for how long? Estimate when it is 1.5 meters up.
- From Figure 2, estimate the widths that give an area of 60 square meters, and describe the rectangle.
More situations
- A club sells mugs. Its profit P, in dollars, when it charges p dollars a mug is modeled by P(p) = -2(p - 4)(p - 20) for 0 \le p \le 20 (a made-up model). Give the sign of a, the vertical intercept, the zeros, the axis, the vertex and the range, each with its meaning.
- Two numbers add to 10. Write their product P as a function of one of them, x. Give the vertex, the domain and the range. Why is the domain not trimmed the way the width in Example 4 is?
- A delivery van’s fuel use F, in liters per 100 km, at a steady v km/h is modeled by F(v) = 0.002(v - 70)^2 + 7 for 30 \le v \le 120 (a made-up model). What does the sign of a tell you? Give the lowest point and its meaning, the fuel use at each end of the domain, and the range. What does it mean that F has no horizontal intercepts?
Reasoning
- The bare formula h(t) = -5t^2 + 8t - 1.4 has range h \le 1.8. Is that the range of the dolphin’s leap? Say what is right about it and what is missing.
- In Keturah’s A(w) = w(18 - w), explain why the widths w and 18 - w always give the same area, and what that says about the axis of symmetry.
- The shells of Examples 2 and 5 both have a = -5. Explain why they can still reach different heights.
- Explain why no rule with a > 0 could model a dolphin’s leap from the start of its run to its return to the water.
Answers
- Opens downward (a = -1). Horizontal intercepts x = 2 and x = 6; vertical intercept f(0) = -(-2)(-6) = -12. Axis x = 4; vertex f(4) = -(2)(-2) = 4, so (4, 4), a maximum. Range y \le 4.
- Vertex (3, -4), a lowest point, since the square is never negative. Vertical intercept p(0) = 9 - 4 = 5. x = 1 is 2 left of the axis x = 3, so the other is x = 5; check: p(5) = 4 - 4 = 0.
- h(0) = -5(-0.3)(-1.1) = -1.65: it starts 1.65 m deep. Out from 0.3 s to 1.1 s, so for 0.8 s. Highest midway: h(0.7) = -5(0.4)(-0.4) = 0.8 m at 0.7 s. Domain 0 \le t \le 1.1, range -1.65 \le h \le 0.8.
- Half a second after its run starts, the dolphin is 1.35 meters above the surface. Check: h(0.5) = -1.25 + 4 - 1.4 = 1.35.
- Launch height 1 m. Equal heights pair up around t = 1.5, so the peak is 12.25 m at 1.5 s. Back at 1 m at t = 3 s.
- The axis is t = 1.5 and 3.07 is 1.57 to its right, so the other zero is 1.5 - 1.57 = -0.07, before launch, meaningless. At t \approx 3.07 s the shell lands (just after h(3) = 1). Domain 0 \le t \le 3.07 (about), range 0 \le h \le 12.25.
- f(3) = f(5) = -1, so the axis is x = 4 and the vertex (4, -2) is a minimum, since the values fall then rise. By symmetry f(6) = f(2) = 2 and f(7) = f(1) = 7. The sign changes between f(2) = 2 and f(3) = -1, and between f(5) = -1 and f(6) = 2: intercepts between 2 and 3 and between 5 and 6.
- Above the surface for 0.2 < t < 1.4, which is 1.2 s. At 1.5 m at about t = 0.55 s and t = 1.05 s, each about 0.25 from the axis. Check: h(0.55) \approx 1.49.
- The level 60 meets the arch at about w = 4.5 and w = 13.5, each about 4.5 from the axis. Both give one rectangle, about 4.5 m by 13.5 m. Check: 4.5 \times 13.5 = 60.75, close to 60.
- a = -2 < 0: as the price rises, profit rises, then falls. P(0) = -2(-4)(-20) = -160: giving the mugs away loses 160 dollars. Zeros p = 4 and p = 20: at these prices the club breaks even. Axis p = 12; vertex P(12) = -2(8)(-8) = 128, so (12, 128): a price of 12 dollars earns the most, 128 dollars. Range -160 \le P \le 128.
- P(x) = x(10 - x), zeros 0 and 10, vertex (5, 25). Domain all real numbers, range P \le 25. Any two numbers may add to 10, negatives too (-2 and 12 give -24); a width must be positive and leave a positive length.
- a = 0.002 > 0: as the speed rises, fuel use falls, then rises. Lowest point (70, 7): the van is thriftiest at 70 km/h, using 7 liters per 100 km. F(30) = 0.002(1600) + 7 = 10.2 and F(120) = 0.002(2500) + 7 = 12. Range 7 \le F \le 12, the top at the end farther from the axis. No horizontal intercepts: at no speed does the van run on no fuel.
- Right: the highest height is 1.8 m, at t = 0.8, inside the domain. Missing: the leap runs only for 0 \le t \le 1.4, and its lowest height is at the start, h(0) = -1.4. The range is -1.4 \le h \le 1.8; h \le 1.8 is the bare formula’s range, a dolphin sinking without end.
- Since 18 - (18 - w) = w, the area at width 18 - w is (18 - w)w = A(w): swapping width and length gives the same rectangle turned. w and 18 - w average to 9, so they sit equally far either side of w = 9, which is therefore the axis: the square, whose width equals its length.
- a fixes only how sharply the arch bends. Parabolas with the same a are slides of each other, so the vertex can sit anywhere: Example 2 peaks at 62.75 m after 3.5 s, Example 5 at 80 m after 4 s.
- With a > 0 the graph opens upward: the height would fall to a lowest point, then rise, with no highest point except at the ends of the domain. A leap rises to a top and falls back, a highest point inside the domain, which needs a < 0.