Modelling with Quadratics
Profit, height, area — whenever one quantity is built by multiplying two linear pieces together, a quadratic appears. The vertex then answers the question a picture can't: the biggest, the smallest, or the exact moment something happens.
A quadratic doesn't usually turn up by itself — it's built. Whenever a real quantity is the product of two linear expressions, the result is quadratic: revenue is price times quantity sold, and the quantity sold often falls linearly as price rises; the area of a rectangular field is length times width, and a fixed length of fencing ties those two together; the height of something thrown or launched follows $s=ut+\tfrac12at^2$, a quadratic in time. Modelling with quadratics is the skill of turning a sentence like this into an equation, then using everything you already know about parabolas to answer the question that was actually asked.
The algebra doesn't change from the last lesson — you'll still complete the square, still use the discriminant, still solve for roots. What's new is the context: every letter stands for something physical, the domain is often restricted to values that make sense, and the answer has to be interpreted, not just stated.
Every modelling question follows the same shape: define the variable in words (what does $x$ actually represent?), write the quadratic expression for the quantity you care about, then complete the square to read off the vertex. That one algebraic step usually answers the question in front of you — a maximum profit, a maximum height, a maximum area — directly.
Reading off the maximum or minimum
Writing a quadratic in vertex form $y=a(x-p)^2+q$ makes the extreme value visible at a glance. Because $(x-p)^2 \geq 0$ for every real $x$, the term $a(x-p)^2$ contributes as little as possible — exactly zero — when $x=p$. So if $a<0$ (the parabola opens downward), $y$ is at its largest when $x=p$, and that largest value is $q$. If $a>0$, the same reasoning gives a smallest value $q$ at $x=p$. Either way, the vertex $(p,\,q)$ is the answer to "what's the best this quantity can be, and when?"
$a(x-p)^2$ is a squared term scaled by $a$: it's the only part of $a(x-p)^2+q$ that changes as $x$ varies, and a square is never negative. So moving $x$ away from $p$ can only push $a(x-p)^2$ further from zero — making $y$ smaller if $a<0$, or larger if $a>0$. The single value of $x$ that avoids this penalty entirely is $x=p$, which is exactly why the vertex is where the maximum or minimum sits.
Restricting the domain to what makes sense
A price can't be negative. A length of fencing can't be longer than the fencing you have. A time before the object was launched doesn't exist. Real models come with a domain — the set of $x$-values that actually mean something — and it's often narrower than "all real numbers".
The vertex is only the answer if it falls within the domain the model is valid for. If a calculation gives a vertex outside the sensible range (say, a negative price), the quantity is instead increasing or decreasing throughout the whole valid domain, and the true best value sits at a boundary of that domain instead. Always state the domain, and glance at whether $x=p$ actually lies inside it, before quoting the vertex as your final answer.
Worked example 1
Two numbers are expected here, not one: the price and the profit it produces. It's easy to complete the square correctly and then only report $x=15$, or only report $200$ — read the question again before you finish, and answer exactly what was asked.
Worked example 2
"Maximum height" is a vertex question — read it straight from the completed-square form. "When does it land" is a roots question — set $h(t)=0$ and solve. The same completed square feeds both, but only because $(t-2)^2=23/5$ was one algebraic step away once the square was already done. Notice too that the quadratic formula would have given the identical surd — completing the square and the formula are the same method wearing different clothes.
Worked example 3
Nothing here was quadratic to begin with — the fencing constraint is linear ($2x + \text{(third side)} = 40$), and area is just length times width. Multiplying two linear expressions is precisely what generates a quadratic, and it's the reason "fixed perimeter, maximum area" problems are one of the most common quadratic-modelling contexts you'll meet.
Finding the roots (where the model equals zero) answers a different question from finding the vertex (the best value). Worked example 2 needed both, for two separate parts of the question — don't reach for the discriminant or the quadratic formula out of habit when the question is really asking "what's the maximum", and don't complete the square when the question is really asking "when does it hit zero". Read what's being asked before choosing the tool.
Check yourself
A company's daily revenue, in pounds, from selling $n$ items is modelled by $R(n) = -3n^2 + 90n$ for $n \geq 0$. Find the number of items that maximises revenue, and state the maximum revenue.
(Hint: factor the $-3$ out of the $n$-terms before completing the square.)
Notice there's no constant term to carry through here — $R(n)$ has no "$+c$", so the completed square simplifies a little faster than in the earlier examples. The method doesn't change either way.