The vertex is the most or the least — turning fences, prices, and fountains into quadratics to optimize.
Because a parabola turns exactly once, its vertex is a very special place: it is the single lowest point of an upward bowl, or the single highest point of a downward one. Everywhere else the curve is on its way up or on its way down — only at the vertex does it pause and turn. That one property makes the quadratic the natural model for every "as much as possible" or "as little as possible" question you will ever meet: the largest area you can fence in, the price that earns the most profit, the peak of a fountain's arc. In this lesson we read the extreme value straight off the vertex, learn what to do when the input is penned into a limited interval, and then walk the full modeling workflow that turns a word problem into a parabola you can solve.
You already know how to find the vertex of y = ax2 + bx + c — from 23.3 Vertex, Axis & Opening, or by completing the square: its x-coordinate is the axis of symmetry x = −b/2a, and its y-coordinate is the height there. The vertex is also the turning point, so it is the curve's most extreme value — but which extreme depends on which way the bowl opens.
If a > 0 the parabola opens upward: the arms rise forever on both sides, so the vertex is the lowest point — a minimum. If a < 0 the parabola opens downward: the arms fall forever, so the vertex is the highest point — a maximum. Either way the extreme value is k, the y of the vertex, and it is reached at x = h = −b/2a.
Find the extreme value of y = x2 − 4x + 7. Complete the square: y = (x2 − 4x + 4) − 4 + 7 = (x − 2)2 + 3. The bracket is never negative, so the smallest y is when (x − 2)2 = 0, that is at x = 2. Since a = 1 > 0, the vertex is a minimum: the least value of y is 3, reached at x = 2.
For y = ax2 + bx + c, the extreme value sits at the vertex (h, k) with h = −b/2a and k = f(h). If a > 0 it is a minimum (smallest y = k); if a < 0 it is a maximum (largest y = k). The h tells you where; the k tells you how much.
In the wide-open world, x can be anything, and the vertex always wins. But real problems often pen x into an interval: a fence side cannot be negative, a price has a floor, a stopwatch runs only so long. When x is restricted to [m, n], the rule changes — you must look at which part of the parabola lives over that interval.
Here is the whole decision in one question: is the axis h inside [m, n]?
Take y = x2 − 4x + 7 again, with axis h = 2 and vertex (2, 3).
On [0, 3]: the axis x = 2 is inside, so the vertex is reachable. Minimum = 3 at x = 2. The farther endpoint from 2 is x = 0 (distance 2 vs. 1), so the maximum is f(0) = 7.
On [0, 1]: the axis x = 2 is outside to the right. For a > 0 the curve is decreasing left of the axis, so on [0, 1] it falls. The maximum is at the left endpoint f(0) = 7; the minimum is at the right endpoint f(1) = 1 − 4 + 7 = 4.
Do not blindly plug in the vertex when it lies outside the allowed interval. The value y = 3 at x = 2 is real, but if x is only allowed up to 1, you can never reach it — the answer there is f(1) = 4, not 3. Always check whether h is between m and n first.
Here is the most famous optimization story of all. You have a fixed length of fence and you want to enclose the largest possible area. There are many shapes you could make — long and thin, or short and squat — and somewhere in between is the best one. A quadratic finds it.
Suppose you have 40 m of fence and you build three sides of a rectangle against a long wall (the wall is the fourth side, for free). Let the two sides perpendicular to the wall each be x, and the side parallel to the wall be y. The fence covers those three sides, so:
2x + y = 40, which gives y = 40 − 2x.
The area is length × width:
A = x · y = x(40 − 2x) = −2x2 + 40x.
That is a quadratic in x with a = −2 < 0, so it opens downward and has a maximum. The axis is at x = −b/2a = −40/(2·(−2)) = 10. So the best side is x = 10 m, the parallel side is y = 40 − 2(10) = 20 m, and the largest area is A = −2(10)2 + 40(10) = 200 m2. The sensible domain is 0 < x < 20 (each side must be a positive length, and y = 40 − 2x must stay positive).
To maximize (or minimize) an area, write the area as a function of one variable, using the fixed total to eliminate the other. Then it is just a quadratic — find the vertex and read off the answer. The most area against a wall with this setup is a rectangle twice as wide as it is deep (y = 2x).
Money problems hide quadratics too. Raise a price and you earn more on each sale — but you also sell fewer. Revenue is price × quantity, and when both move in opposite directions, their product is a parabola with a peak.
A shop sells 100 items a week at $10 each. Market research says every $1 rise in price loses 5 sales. Let x be the number of dollars added to the price. Then:
price = (10 + x) dollars, quantity = (100 − 5x) items.
Weekly revenue is their product:
R = (10 + x)(100 − 5x) = 1000 + 50x − 5x2 = −5x2 + 50x + 1000.
This opens downward (a = −5), so it has a maximum at x = −b/2a = −50/(2·(−5)) = 5. The best move is to add $5: the price becomes $15, quantity is 100 − 5(5) = 75 items, and revenue peaks at R = 15 × 75 = $1125 — up from the original 10 × 100 = $1000. The sensible domain is 0 ≤ x ≤ 20 (beyond x = 20 the quantity 100 − 5x would go negative — you would be selling nothing).
Throw a ball, spray a fountain, launch a basketball toward the hoop — under gravity, the height traces a parabola. Because gravity is constant, the height is a quadratic in time, and that means its peak is exactly the vertex.
Suppose a ball is thrown straight up so that its height in metres after x seconds is:
y = −5x2 + 20x.
It opens downward (a = −5), so the highest point is the vertex, at x = −b/2a = −20/(2·(−5)) = 2. At that instant the peak height is y = −5(2)2 + 20(2) = −20 + 40 = 20 m. When does it land? Set y = 0: −5x2 + 20x = −5x(x − 4) = 0, so x = 0 (the launch) or x = 4 s (the landing). Notice the launch and the landing are symmetric about the peak: 0 and 4 average to 2, the vertex's x.
Keep the two vertex numbers straight. The x at the vertex is when the peak happens (t = 2 s); the k is how high it gets (20 m). A question asking "how long until the peak?" wants h; "how high does it go?" wants k. Reading the wrong one off the vertex is the classic slip.
Every example in this lesson followed the same five-step path. Once you can name the steps, any new "biggest" or "smallest" problem becomes routine: translate it into a quadratic, find the vertex, and answer.
| Step | What you do | Fence example |
|---|---|---|
| 1 | Name the variable | let x = a side (m) |
| 2 | Write the quantity as a function of it | A = x(40 − 2x) |
| 3 | Fix the realistic range (domain) | 0 < x < 20 |
| 4 | Find the vertex (check it lies in range; else use an endpoint) | x = 10, inside ✓ |
| 5 | Answer the actual question, with units | 200 m2 |
The five steps — (1) name, (2) write the function, (3) fix the domain, (4) find the vertex (or endpoint), (5) answer with units — are the whole game. Step 4 is where the quadratic's single turning point earns its keep; step 3 is the one most people skip, and it is exactly what decides whether the vertex or an endpoint is the true answer.
A parabola turns once, and that turning point — the vertex (h, k) — is the most or the least the function ever gets. When a > 0 it is a minimum; when a < 0 it is a maximum; the value is always k, reached at x = −b/2a. If the input is confined to an interval [m, n], first check whether the axis is inside: if so the vertex competes with the farther endpoint, and if not the extremes are both at the endpoints. Real problems — the largest fenced area, the price for top revenue, the peak of a thrown ball — all become quadratics through one habit: write the quantity as a function of a single variable, fix its domain, then find the vertex. That is optimization, and the parabola makes it visible. This completes the quadratic's three faces — function, equation, and inequality, tied together in 23.5; next, in 23.7 Power Functions, we lift y = x2 to the whole power family y = xα.
Six questions to lock it in. Tap the answer you think is right.
This lesson is the payoff of the stage: it puts the vertex to work as an optimizer and walks a repeatable modeling routine. It aligns with F-IF.B.4 (interpret a maximum or minimum in context), F-IF.C.8a (use the vertex form to read off the max or min), A-CED.A.2 and A-SSE.B.3b (model a situation and rewrite to optimize), and F-BF.A.1 (build a function describing a relationship).
The three widgets are the heart of it. Slide the window dramatizes interval extremes — students watch the minimum hop from an endpoint to the vertex the moment the axis x = 2 slips inside. The fence lets them search by hand and discover the peak that calculus would later confirm. The clock separates when the peak occurs (h) from how high it is (k).
Three misconceptions to watch for: (1) plugging the vertex into the answer even when it lies outside the allowed domain — on a limited interval, always check whether h is between the endpoints first; (2) forgetting the realistic range or the units — "x = 10" is not an answer; "200 m²" is; (3) confusing when the peak happens (h) with how high it is (k). A good habit: after finding the vertex, re-read the question and underline exactly what it asks for before writing the final line.