How To Find The Maximum Of A Quadratic Function

You’re staring at a parabola on a graph, or maybe just an equation on a page: f(x) = -2x² + 8x - 5*. A peak. Somewhere along that curve, there’s a highest point. The maximum.

If the coefficient on that x² term is negative, the arms of the parabola point down. Because of that, if it’s positive, the arms point up — you’ve got a minimum instead, and no maximum at all (unless you restrict the domain, but we’ll get to that). That means a maximum exists. Most students freeze right here. They memorize a formula, plug in numbers, and hope the answer matches the back of the book.

There’s a better way. Plus, actually, there are a few. And understanding why they work beats memorizing what* to type into a calculator every single time.

What Is a Quadratic Function (and When Does It Have a Maximum?)

A quadratic function is any function you can write in the form f(x) = ax² + bx + c*, where a, b, and c are real numbers and a isn't zero. That “a ≠ 0” part matters. If a were zero, the x² term vanishes and you’re left with a line. Lines don’t have peaks or valleys unless you chop them off at the ends.

The graph of a quadratic is a parabola. Symmetric. Smooth. U-shaped — or ∩-shaped.

Here’s the key: the sign of a tells you everything about the vertex.

• a > 0: Parabola opens upward. And vertex is the minimum. - a < 0: Parabola opens downward. Vertex is the maximum.

That’s it. If a is negative, the vertex is the maximum value of the function. The y-coordinate of the vertex is the maximum output. The x-coordinate tells you where* it happens.

The Three Forms You’ll See

Textbooks love showing you three forms. They’re not just busywork — each one reveals something different.

Standard form: f(x) = ax² + bx + c* Good for: identifying a, b, c quickly. Plugging into the vertex formula. Finding the y-intercept (it’s just c).

Vertex form: f(x) = a(x - h)² + k* Good for: reading the vertex instantly. It’s (h, k). No calculation required. If you have this form and a is negative, k is your maximum. Done.

Factored form: f(x) = a(x - r₁)(x - r₂)* Good for: finding the x-intercepts (roots) r₁ and r₂. The vertex sits exactly halfway between them because of symmetry. The x-coordinate of the vertex is the average of the roots: (r₁ + r₂) / 2.

You can convert between these forms. Factoring (or the quadratic formula) turns standard into factored. Which means completing the square turns standard into vertex. The algebra gets messy sometimes, but the geometry stays the same.

Why Finding the Maximum Matters

This isn’t just a homework exercise. Optimization — finding the best possible outcome under constraints — is one of the main reasons calculus and algebra exist in the real world.

Business and economics. Revenue, profit, and cost functions are often modeled as quadratics (at least locally). A company selling a product might find that profit P(x) = -5x² + 300x - 2000*, where x is the number of units sold. The maximum of that parabola? That’s the production level that maximizes profit. The y-value? That’s the maximum profit itself. Getting this wrong means leaving money on the table — or worse, losing money by overproducing.

Physics and engineering. Projectile motion. Throw a ball, launch a rocket, shoot an arrow. Ignoring air resistance, the height h(t)* as a function of time t is quadratic: h(t) = -½gt² + v₀t + h₀*. The coefficient on t² is negative (gravity pulls down). The vertex gives you the maximum height and the exact time it occurs. Civil engineers use the same math for arch bridges and parabolic reflectors — the peak of the arch or the focus of the reflector depends on that vertex.

Geometry problems. “You have 100 meters of fencing. Build a rectangular pen against a barn (so you only need three sides). What dimensions maximize the area?” Area A = x(100 - 2x) = -2x² + 100x*. That’s a downward parabola. The maximum area happens at the vertex. This exact problem shows up in calculus textbooks as a “first optimization problem,” but you don’t need calculus to solve it. Algebra gets you there.

How to Find the Maximum: Three Reliable Methods

You have options. The “best” method depends on what form the function is in, what tools you’re allowed to use, and how much time you have.

Method 1: The Vertex Formula (The Shortcut)

If you have standard form f(x) = ax² + bx + c*, the x-coordinate of the vertex is always:

x = -b / (2a)

Yes, always. It comes from completing the square (see Method 2), but you don’t need to re-derive it every time. Memorize it

, understand where it comes from, and move on.

Plug that x back into the original function to get the y-value (the maximum or minimum, depending on the sign of a).

Example. f(x) = -2x² + 12x - 7*. Here a = -2*, b = 12*. So x = -12 / (2 × -2) = -12 / -4 = 3*. Plug in: f(3) = -2(9) + 12(3) - 7 = -18 + 36 - 7 = 11*. The vertex is (3, 11). Since a is negative, it’s a maximum. The function never exceeds 11.

This method is fast, reliable, and works for any standard-form quadratic. Day to day, no factoring required, no square roots, no messy algebra. If you remember one formula from this whole article, make it this one.

Method 2: Completing the Square (The Deep Understanding)

This is where vertex form comes from. It’s more work than Method 1, but it reveals why the vertex formula works, and it’s essential for more advanced topics like conic sections and integration.

Start with f(x) = ax² + bx + c*. Factor a out of the first two terms:

For more on this topic, read our article on how many minutes are in 6 hours or check out what is 3 8 in decimal form.

f(x) = a(x² + (b/a)x) + c*

Now focus on the expression inside the parentheses. You want to turn x² + (b/a)x* into a perfect square. So the rule: take half the coefficient of x, square it, and add it. Because of that, half of (b/a) is (b/2a), and squaring gives b²/4a²*. But you have to add this inside* the parentheses, which means you’re actually adding a × b²/4a² = b²/4a* to the whole function. To keep the function equal, subtract that same amount outside.

f(x) = a(x² + (b/a)x + b²/4a²) + c - b²/4a*

The expression inside the parentheses is now a perfect square: (x + b/2a)². So:

f(x) = a(x + b/2a)² + (c - b²/4a)*

That’s vertex form. The vertex is at (-b/2a, c - b²/4a). Notice the x-coordinate is the same as Method 1: -b/2a. The y-coordinate simplifies (with a little algebra) to (4ac - b²)/4a, which is the value you get by plugging x = -b/2a* back into the original function.

Completing the square looks intimidating the first few times, but it’s a mechanical process. Do it enough and it becomes second nature — and you’ll start seeing quadratic structure everywhere.

Method 3: Using the Roots (When You Have Them)

If you already know the x-intercepts r₁ and r₂, you don’t need the vertex formula. The axis of symmetry is the vertical line halfway between the roots. So the x-coordinate of the vertex is just the average:

x = (r₁ + r₂) / 2

Then plug that x into the function to get the maximum (or minimum) value.

This method is common in physics and engineering problems where you know when something starts and ends — for example, the time a projectile leaves the ground and the time it lands. The maximum height occurs at the midpoint of that interval.

Worked Example: The Fence Problem

Let’s solve the classic optimization problem from earlier using all three methods, so you can see how they connect.

Problem. You have 100 meters of fencing. One side of a rectangular pen will be against a barn, so you only need fencing for three sides. Find the dimensions that maximize the area.

Step 1: Set up the function. Let x be the width of the pen (the two sides perpendicular to the barn). The side parallel to the barn is 100 - 2x. Area:

A(x) = x(100 - 2x) = -2x² + 100x*

We want to maximize A(x)* for x between 0 and 50 (you can’t use negative fencing or more than 100 meters total).

Method 1: Vertex formula. Here a = -2*, b = 100*.

x = -100 / (2 × -2) = -100 / -4 = 25*

A(25) = -2(625) + 100(25) = -1250 + 2500 = 1250*

Maximum area is 1250 square meters, achieved when the pen is 25 meters wide (the two fenced sides) and 50 meters long (the side parallel to the barn).

Method 2: Completing the square.

A(x) = -2x² + 100x* A(x) = -2(x² - 50x)* A(x) = -2(x² - 50x + 625 - 625)* A(x) = -2((x - 25)² - 625)* A(x) = -2(x - 25)² + 1250*

Vertex form: A(x) = -2(x - 25)² + 1250*. Because of that, the vertex is (25, 1250). Same answer.

Method 3: Using the roots. The roots of A(x) = -

The roots of (A(x) = -2x^{2}+100x) are obtained by solving

[ -2x^{2}+100x = 0 ;\Longrightarrow; -2x(x-50)=0, ]

so the two intercepts are (x_{1}=0) and (x_{2}=50).
The axis of symmetry – and therefore the (x)-coordinate of the vertex – is the midpoint of these roots:

[ x_{\text{vertex}} = \frac{x_{1}+x_{2}}{2}= \frac{0+50}{2}=25. ]

Plugging this back into the area function gives the maximum area:

[ A(25)= -2(25)^{2}+100(25)= -1250+2500 = 1250\ \text{m}^{2}. ]

Thus the pen should be 25 m wide (the two fenced sides) and 50 m long (the side against the barn), yielding a maximal area of 1250 m².

Bringing It All Together

All three approaches lead to the same dimensions and area, which is a reassuring check on any calculation:

Each method has its own strengths. g.The vertex formula is quickest when the coefficients are known, completing the square reveals the parabola’s shape and is useful for deeper algebraic manipulation, and the root‑average technique shines when the intercepts are already at hand (e., in physics problems describing start‑ and end‑times).

Final Takeaway

Whether you’re optimizing a garden layout, analyzing projectile motion, or simply trying to understand the geometry of a parabola, you now have three reliable tools to locate its vertex. But mastering all of them gives you flexibility: choose the one that fits the information you have and the level of insight you need. In the fence problem, they all converge on the same optimal design—25 m of width and 50 m of length—for a perfectly maximized area of 1250 m².