Maxima and Minima of Functions

What this move does

With no restriction, the maximum or minimum of a quadratic function is at the vertex. But problems often let $x$ take values only in an interval, and then the vertex may not even be there.

y = a (x - h)^{2} + k, m \leq x \leq n

Complete the square to find the axis of symmetry $x = h$ .
See where the axis is relative to the interval $[m, n]$ : to the left, inside or to the right.
If it opens upward: in those three cases the minimum is at the left endpoint, the vertex or the right endpoint; the maximum is at the endpoint farther from the axis. (If it opens downward, swap “maximum” and “minimum”.)

In other words, there are only three candidates: the two endpoints, plus the vertex if it lies in the interval.

Predict first

$y = x^{2} - 4 x + 5$ , with $x$ allowed only in $3 \leq x \leq 5$ . What is the minimum value of $y$ ?

Choose an answer first; then you can see the result.

Work through one: moving axis, fixed interval

The interval stays put; the axis, with a letter in it, moves around.

Example: max and min of f(x) = (x − h)² − 2 on −1 ≤ x ≤ 2

Step 1 / 5

Axis to the left of the interval
$h = - 3$ . The piece of graph over the interval lies entirely to the right of the axis and keeps rising: the minimum is at the left endpoint and the maximum at the right endpoint.

$h < - 1 : min = f (- 1) = (h + 1)^{2} - 2$
Axis inside the interval
Axis past the midpoint of the interval
Axis to the right of the interval
Put the three cases together

Once more: fixed axis, moving interval

This time the parabola stays put and the interval, with a letter in it, slides. The idea is exactly the same: see which side of the interval the axis falls on.

Example: minimum of f(x) = x² − 2x − 1 on t ≤ x ≤ t + 2

Step 1 / 5

Complete the square first
$f (x) = (x - 1)^{2} - 2$ , so the axis $x = 1$ is fixed. The interval $[t, t + 2]$ has width $2$ .

$f (x) = (x - 1)^{2} - 2$
Interval entirely left of the axis
Interval covering the axis
Interval entirely right of the axis
Write the answer in pieces

One problem, several solutions

Find the maximum and minimum of $y = x^{2} - 6 x + 10$ on $1 \leq x \leq 4$ .

Complete the square, look at the axis

$y = (x - 3)^{2} + 1$ . The axis $x = 3$ is inside the interval, so the minimum is $1$ .

The midpoint of the interval is $2.5$ . The axis is to the right of the midpoint, so the left endpoint $x = 1$ is farther from it: the maximum is $f (1) = 5$ .

CostShows you why; the only way when there are letters.

Work out all three candidates

The candidates: left endpoint, right endpoint, and the vertex inside the interval.

$f (1) = 5$ , $f (4) = 2$ , $f (3) = 1$ .

The largest is $5$ and the smallest is $1$ .

CostNo thinking needed and quick with numbers; with letters you cannot compare the sizes.

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: If the parabola opens upward, the minimum is the vertex's y-coordinate

When the axis is inside the interval, the vertex is the lowest point of that piece. No problem.

Practice

Exercise 1Basic

Find the maximum of $y = x^{2} - 2 x + 3$ on $0 \leq x \leq 3$ .

Exercise 2Basic

Find the maximum of $y = - x^{2} + 4 x + 1$ on $3 \leq x \leq 5$ .

Exercise 3Advanced

Real numbers $x$ and $y$ satisfy $x + y = 6$ . Find the maximum of $x y$ .

Exercise 4Advanced

The minimum of $f (x) = x^{2} - 2 a x + 1$ on $0 \leq x \leq 2$ is $- 8$ . Find $a$ .

Exercise 5Challenge

The minimum of $f (x) = x^{2} - 2 x + 2$ on $t \leq x \leq t + 1$ is $5$ . Find all possible values of $t$ .

What this move does

Work through one: moving axis, fixed interval

Example: max and min of f(x) = (x − h)² − 2 on −1 ≤ x ≤ 2

Once more: fixed axis, moving interval

Example: minimum of f(x) = x² − 2x − 1 on t ≤ x ≤ t + 2

Complete the square, look at the axis

Work out all three candidates

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Taking the vertex without looking at the interval

An axis with a letter in it, and no case analysis