Algebra 1 · A.SSE.B.3B · F.IF.C.7A

Maximum, minimum and range of a quadratic function

A parabola turns around exactly once. What does that one turning point tell you about every output the function can give?

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Average rate of change from a formulaAlgebra 1 Turning points, maximums and minimumsAlgebra 1

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Quadratic functions in contextAlgebra 1

Where you are 3 in Quadratic functions

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Maximum, minimum and range of a quadratic function

Quadratic functions in context Reading a quadratic in its three forms

When you reach for it

A quadratic function f(x) = ax^2 + bx + c, with a \ne 0, has a parabola for its graph, and a parabola turns around exactly once, at its vertex. Reach for this lesson when a question about outputs is really about that point: the smallest or largest output and where it occurs, the range, where the graph falls and rises, the outputs from a limited set of inputs, and how fast the function changes when, unlike a line, its rate keeps changing.

The procedure

The vertex gives the least or greatest output

Every quadratic can be written in vertex form

f(x) = a(x - h)^2 + k

where (h, k) is the vertex. Finding the vertex is its own lesson; the question here is what it says about outputs.

The square (x - h)^2 is never negative, and it is 0 only when x = h. Suppose a > 0. Then a(x - h)^2 \ge 0 for every x, so

f(x) = a(x - h)^2 + k \ge k

with equality only at x = h. So k is the minimum value of f: no input gives less, and only x = h gives exactly k. If a < 0, a non-negative square times a negative number is never positive, so a(x - h)^2 \le 0, f(x) \le k, and k is the maximum value.

Keep two things apart. The value is an output: k, the vertex’s second coordinate. Where it occurs is an input: h, the first coordinate. “The minimum is -5” and “the minimum occurs at x = 2” answer different questions.

This minimum or maximum is absolute: least or greatest over every real input, not only nearby ones. A relative maximum or minimum only has to beat its neighbors, and a general function may have several. A quadratic has one turning point, so its relative extreme is also its absolute one. The other kind does not exist at all: an upward parabola has no maximum, because a(x - h)^2 grows past any number as x moves far enough from h.

Finding it from each form

Standard form ax^2 + bx + c. The vertex lies on the axis \displaystyle x = -\frac{b}{2a} (derived there). Substitute that input into f to get the value.

Vertex form a(x - h)^2 + k. Read it off: the value is k, at x = h. Mind the sign inside the brackets: (x + 5)^2 = (x - (-5))^2, so h = -5.

Factored form a(x - r)(x - s). The zeros r and s are mirror images in the axis, so the axis lies midway between them, at \displaystyle x = \frac{r + s}{2}. Substitute that input to get the value.

In every form the sign of a decides which kind: a > 0 gives a minimum, a < 0 a maximum. In factored form a is the number in front of the brackets, since multiplying out gives ax^2 first.

Domain and range

Any real number can be squared, so the domain of a quadratic is all real numbers, (-\infty, \infty). The range is every number on one side of k:

\begin{aligned} a > 0&:\ \text{range } [k, \infty) \\ a < 0&:\ \text{range } (-\infty, k] \end{aligned}

or, as inequalities, y \ge k and y \le k. In Interval and set-builder notation the bracket at k is square because k is reached, at x = h; the end at infinity is always round, because infinity is not a number and is never reached.

Every number past k really is an output. For a > 0 and any y > k, solve a(x - h)^2 + k = y: then \displaystyle (x - h)^2 = \frac{y - k}{a}, which is positive, so

x = h \pm \sqrt{\frac{y - k}{a}}

The same line shows more: every output except k is reached at exactly two inputs, one on each side of the axis, and k at one.

Where the graph falls and rises

Take a > 0. Right of the axis, as x increases past h, the distance x - h grows, so its square grows, so f(x) grows: f is increasing on (h, \infty). Left of the axis, as x increases toward h, the distance h - x shrinks, so the square shrinks: f is decreasing on (-\infty, h). For a < 0 multiplying by the negative a flips both: increasing on (-\infty, h), decreasing on (h, \infty). The general idea of rising and falling stretches is its own lesson; for a parabola the axis is the only place the direction changes.

Figure 1The vertex is the lowest point, so every output is at least -5. Left of the axis the graph falls; right of it, it rises.

On a restricted set of inputs

Sometimes only an interval of inputs s \le x \le t is allowed. The graph is then a piece of the parabola, and its range is decided by at most three outputs: f(s), f(t), and f(h) when the vertex’s input h lies inside the interval. On each side of the axis the function moves only one way, so nothing higher or lower lies between those points; and the graph has no breaks, so every output in between is reached.

How fast a quadratic changes

The average rate of change of f from x = p to x = q is \displaystyle \frac{f(q) - f(p)}{q - p}. For a quadratic it simplifies. The c terms cancel, and the rest factors, q^2 - p^2 by the difference of squares:

\begin{aligned} f(q) - f(p) &= a(q^2 - p^2) + b(q - p) \\ &= (q - p)\big(a(q + p) + b\big) \end{aligned}

Dividing by q - p, the average rate from p to q is

a(p + q) + b

Two facts follow.

Over equal-width intervals the rate changes by a fixed amount. Slide an interval of width w one step right, from [p, p + w] to [p + w, p + 2w]. The sum p + q grows by 2w, so the rate grows by 2aw, every time. A linear function keeps one rate; a quadratic’s rate goes up (or, for a < 0, down) by equal steps. In a table of outputs at equally spaced inputs, the differences between neighboring outputs are rate times w, so they change by 2aw^2 each step: the second differences are constant.

Over an interval centered on the axis the rate is 0. If p and q are mirror images about h, then \displaystyle p + q = 2h = -\frac{b}{a}, and a(p + q) + b = -b + b = 0. The two ends sit at the same height, by symmetry.

Worked examples

Example 1

From standard form

Find the least or greatest value of each function, the input where it occurs, and the range.

(a) f(x) = 2x^2 - 8x + 3 \qquad (b) g(x) = -3x^2 + 4x + 1

Solution

(a) a = 2 is positive, so the parabola opens upward and has a minimum. Its axis is

x = -\frac{-8}{2(2)} = 2

and the value there is f(2) = 2(4) - 16 + 3 = -5. The minimum value is -5, at x = 2, and the range is [-5, \infty). The figure shows the same thing: f falls on (-\infty, 2) and rises on (2, \infty).

(b) a = -3 is negative, so there is a maximum. The axis is \displaystyle x = -\frac{4}{2(-3)} = \frac{2}{3}, and

\begin{aligned} g\left(\tfrac{2}{3}\right) &= -3 \cdot \tfrac{4}{9} + \tfrac{8}{3} + 1 \\ &= -\tfrac{4}{3} + \tfrac{8}{3} + \tfrac{3}{3} = \tfrac{7}{3} \end{aligned}

The maximum value is \displaystyle \frac{7}{3}, at \displaystyle x = \frac{2}{3}, and the range is \displaystyle \left(-\infty, \frac{7}{3}\right].

Example 2

From vertex form

Find the greatest or least value, the range, and where each function increases and decreases.

(a) \displaystyle u(x) = -\frac{1}{2}(x - 3)^2 + 8 \qquad (b) v(x) = 4(x + 5)^2

Solution

(a) Here \displaystyle a = -\frac{1}{2}, h = 3 and k = 8. Since a < 0, the maximum value is 8, at x = 3, and the range is (-\infty, 8]. A downward parabola rises to its vertex and then falls, so u is increasing on (-\infty, 3) and decreasing on (3, \infty). Spot check: u(1) = u(5) = 6, both below 8.

(b) Write it as 4(x - (-5))^2 + 0: h = -5, k = 0, a = 4 > 0. The minimum value is 0, at x = -5, and the range is [0, \infty). v is decreasing on (-\infty, -5) and increasing on (-5, \infty).

Example 3

From factored form

Find the greatest or least value of each function, where it occurs, and the range.

(a) f(x) = -2(x - 2)(x - 10) \qquad (b) g(x) = 3x(x + 6)

Solution

(a) The zeros are 2 and 10, so the axis is midway, at \displaystyle x = \frac{2 + 10}{2} = 6. Then

f(6) = -2(6 - 2)(6 - 10) = -2(4)(-4) = 32

The number in front is -2, so the parabola opens downward: the maximum value is 32, at x = 6, and the range is (-\infty, 32].

(b) Read 3x(x + 6) as 3(x - 0)(x - (-6)): the zeros are 0 and -6, and the axis is at x = -3. Then g(-3) = 3(-3)(3) = -27. The number in front is 3 > 0, so the minimum value is -27, at x = -3, and the range is [-27, \infty). Nothing had to be multiplied out.

Example 4

On a restricted domain

Find the range of f(x) = x^2 - 6x + 3 when (a) -1 \le x \le 4 \qquad (b) 4 \le x \le 6.

Solution

The parabola opens upward, its axis is \displaystyle x = -\frac{-6}{2(1)} = 3, and f(3) = 9 - 18 + 3 = -6. On all real inputs the range would be [-6, \infty).

(a) The input 3 lies inside [-1, 4], so the vertex is part of this piece and gives its least output, -6. The greatest output must come at an end:

\begin{aligned} f(-1) &= 1 + 6 + 3 = 10 \\ f(4) &= 16 - 24 + 3 = -5 \end{aligned}

The larger is 10, so the range is [-6, 10]: the left end lies farther from the axis.

Figure 2With inputs from -1 to 4 only, the outputs run from the vertex’s -6 up to the left end’s 10.

(b) Now 3 is not in [4, 6]. The whole piece lies right of the axis, where an upward parabola only rises, so the least output is at the left end and the greatest at the right: f(4) = -5 and f(6) = 36 - 36 + 3 = 3. The range is [-5, 3]. The vertex’s -6 plays no part, because no allowed input reaches it.

Example 5

How the rate changes

For f(x) = 2x^2 + 4x - 1:

(a) find the average rate of change over each of [-3, -2], [-2, -1], [-1, 0], [0, 1] and [1, 2], and describe the pattern;

(b) find the average rate of change over [-3, 1].

Solution

(a) Work out the outputs first.

x	-3	-2	-1	0	1	2
f(x)	5	-1	-3	-1	5	15

Each interval is 1 wide, so each rate is just the change in output: -6, -2, 2, 6, 10. Each is 4 more than the one before, which is 2aw = 2(2)(1). The formula agrees on any one of them; on [-3, -2], a(p + q) + b = 2(-5) + 4 = -6. The rates are negative left of the axis x = -1 and positive right of it: the falling and rising of the graph, measured.

(b) The axis is \displaystyle x = -\frac{4}{2(2)} = -1, and [-3, 1] runs 2 units either side of it. So the rate is 0:

\frac{f(1) - f(-3)}{1 - (-3)} = \frac{5 - 5}{4} = 0

Example 6

A half-pipe

Gianna’s skateboard half-pipe, seen from the end, has a curved surface modeled by

h(x) = \frac{3}{16}(x - 4)^2, \qquad 0 \le x \le 8

where x is the distance across from one lip, in meters, and h(x) is the height above the lowest point, in meters. (a) Find the range of heights. (b) Find the average rate of change of height over each meter from x = 0 to x = 4, and say what the pattern means.

Solution

(a) \displaystyle a = \frac{3}{16} > 0 and the vertex is (4, 0), with 4 inside [0, 8], so the least height is 0 in the middle. The ends give \displaystyle h(0) = \frac{3}{16}(16) = 3 and h(8) = 3. The range is [0, 3]: the surface runs from the floor of the pipe up to 3 meters at each lip.

(b) The heights at x = 0, 1, 2, 3, 4 are 3, \displaystyle \frac{27}{16}, \displaystyle \frac{12}{16}, \displaystyle \frac{3}{16} and 0. Over each meter the average rates are

-\frac{21}{16},\quad -\frac{15}{16},\quad -\frac{9}{16},\quad -\frac{3}{16}

meters of height per meter across. Each is \displaystyle \frac{6}{16} = \frac{3}{8} more than the last, which is \displaystyle 2aw = 2 \cdot \frac{3}{16} \cdot 1. The surface is steepest at the lip, dropping about 1.3 meters over the first meter, and flattens steadily toward the bottom. Quadratic functions in context reads such models in full.

Boundaries

The function must be quadratic, a \ne 0. With a = 0 it is linear: no vertex, no maximum or minimum unless it is constant, and the range of a non-constant line is all real numbers.

One kind of extreme, never both. On all real inputs an upward parabola has a minimum and no maximum, and a downward one the reverse. The range is always a half-line, never all real numbers.

A restricted domain changes the answer. The vertex counts only if its input is allowed, and the ends of the interval count too. If an end is left out, as in 0 < x < 2, its output is not reached: x^2 on that interval has range (0, 4) and neither a least nor a greatest value.

The rate pattern needs equal widths. The rate grows by 2aw only from one interval to the next of the same width w, and the second differences of a table are constant only when its inputs are equally spaced. The formula a(p + q) + b holds for any interval, but only for a quadratic.

Where this goes next

Word problems asking for the largest area or the highest point of a throw are this lesson applied, in maximum, minimum and projectile problems. Checking the ends as well as the turning point is the start of optimization for functions that turn many times. And the rate pattern sorts families: a line has a constant rate, a quadratic constant second differences, an exponential a constant ratio, as growth compared lays out.

Exercises

Greatest or least value

Say whether each function has a greatest or a least value, then find that value and the input where it occurs.

f(x) = x^2 + 10x + 14
g(x) = -x^2 - 6x + 4
f(x) = 4x^2 - 6x + 1
h(t) = -2t^2 - 10t + 5

From vertex form and factored form

Give the greatest or least value, where it occurs, and the range.

f(x) = 3(x - 2)^2 + 7
g(x) = -(x + 7)^2 - 4
f(x) = (x - 3)(x - 11)
\displaystyle g(x) = -\frac{1}{2}(x + 10)(x - 4)

Domain, range, rising and falling

For f(x) = -2x^2 + 12x - 5, give the domain and range, and the intervals where f increases and decreases.
For \displaystyle g(x) = \frac{1}{2}x^2 - 2x + 1, give the range and the intervals where g increases and decreases.
Write a quadratic in vertex form whose range is (-\infty, 5] and which is increasing exactly on (-\infty, -1).

Restricted domain

Find the range of f(x) = x^2 + 6x + 1 for -4 \le x \le 0.
Find the range of g(x) = -x^2 + 2x + 4 for 2 \le x \le 5.

How fast it changes

For f(x) = 3x^2 + 2x, find the average rate of change over [0, 1], [1, 2] and [2, 3], and say by how much it changes each time.
For f(x) = -x^2 + 10x - 3, find the average rate of change over [2, 8] and over [2, 7].
A radio dish has a surface modeled by \displaystyle d(x) = \frac{1}{40}(x - 10)^2 meters above its lowest point, x meters across, for 0 \le x \le 20. (a) Find the range of heights. (b) Find the average rate of change over [0, 2] and over [2, 4]. (c) Find the average rate of change over [5, 15] without working out any heights.

Reasoning

Explain why a quadratic with a > 0 has no maximum value.
Explain why every output of f(x) = a(x - h)^2 + k other than k comes from exactly two inputs, and why the graph cannot turn around a second time.

Answers

Least value -11, at x = -5. a = 1 > 0; axis \displaystyle x = -\frac{10}{2} = -5; f(-5) = 25 - 50 + 14.
Greatest value 13, at x = -3. a = -1 < 0; axis \displaystyle x = -\frac{-6}{2(-1)} = -3; g(-3) = -9 + 18 + 4.
Least value \displaystyle -\frac{5}{4}, at \displaystyle x = \frac{3}{4}. Axis \displaystyle x = \frac{6}{8} = \frac{3}{4}; \displaystyle f\left(\frac{3}{4}\right) = \frac{9}{4} - \frac{18}{4} + \frac{4}{4}.
Greatest value \displaystyle \frac{35}{2}, at \displaystyle t = -\frac{5}{2}. Axis \displaystyle t = -\frac{-10}{2(-2)} = -\frac{5}{2}; \displaystyle h\left(-\frac{5}{2}\right) = -\frac{25}{2} + 25 + 5.
Least value 7, at x = 2; range [7, \infty). Read off h = 2, k = 7, and a = 3 > 0.
Greatest value -4, at x = -7; range (-\infty, -4]. h = -7, k = -4, and a = -1 < 0.
Least value -16, at x = 7; range [-16, \infty). Axis midway between 3 and 11; f(7) = (4)(-4).
Greatest value \displaystyle \frac{49}{2}, at x = -3; range \displaystyle \left(-\infty, \frac{49}{2}\right]. Zeros -10 and 4, axis x = -3; \displaystyle g(-3) = -\frac{1}{2}(7)(-7), and the number in front is negative.
Domain (-\infty, \infty); range (-\infty, 13]; increasing on (-\infty, 3) and decreasing on (3, \infty). Axis \displaystyle x = -\frac{12}{2(-2)} = 3 and f(3) = -18 + 36 - 5 = 13; it opens downward.
Range [-1, \infty); decreasing on (-\infty, 2) and increasing on (2, \infty). Axis \displaystyle x = -\frac{-2}{2 \cdot \frac{1}{2}} = 2 and g(2) = 2 - 4 + 1 = -1; it opens upward.
For example f(x) = -(x + 1)^2 + 5; any a(x + 1)^2 + 5 with a < 0 works. A range of (-\infty, 5] means a maximum of 5, so k = 5 and a < 0. A downward parabola increases up to its axis and no further, so the axis is x = -1.
[-8, 1]. The axis x = -3 is inside, and f(-3) = 9 - 18 + 1 = -8 is the least output. The ends give f(-4) = -7 and f(0) = 1.
[-11, 4]. The axis x = 1 is outside, and the whole piece lies right of a downward parabola’s axis, where it falls: g(2) = 4 and g(5) = -25 + 10 + 4 = -11.
5, 11 and 17; each is 6 more than the last, 2aw = 2(3)(1). The outputs at 0, 1, 2, 3 are 0, 5, 16, 33.
0 over [2, 8], which is centered on the axis x = 5: f(2) = f(8) = 13. Over [2, 7] it is 1: f(7) = 18, and \displaystyle \frac{18 - 13}{5} = 1. The formula agrees: -(2 + 7) + 10 = 1.
(a) \displaystyle \left[0, \frac{5}{2}\right]: least 0 at the vertex x = 10, and \displaystyle d(0) = d(20) = \frac{100}{40}. (b) \displaystyle -\frac{9}{20} and \displaystyle -\frac{7}{20}, from heights \displaystyle \frac{5}{2}, \displaystyle \frac{8}{5} and \displaystyle \frac{9}{10} at x = 0, 2, 4; they differ by \displaystyle \frac{1}{10} = 2 \cdot \frac{1}{40} \cdot 2. (c) 0, because [5, 15] is centered on the axis x = 10.
Write it as a(x - h)^2 + k with a > 0. Pick any number M as a candidate maximum. Taking x = h + d gives f(x) = ad^2 + k, and choosing d large enough that ad^2 > M - k makes f(x) > M. So no number is the greatest output.
Solving a(x - h)^2 + k = y gives \displaystyle (x - h)^2 = \frac{y - k}{a}. If that is positive there are two inputs, \displaystyle h \pm \sqrt{\frac{y - k}{a}}, one each side of h; if negative, none; it is 0 only for y = k, giving one. No second turn: if h < x_1 < x_2 then (x_1 - h)^2 < (x_2 - h)^2, so for a > 0 the outputs only rise right of the axis (for a < 0, only fall), and the mirror argument works on the left. The axis is the only place the direction changes.