Quadratics show up constantly in real situations that have a "sweet spot" — a highest point, a largest area, a biggest profit. A thrown ball rises and falls, so its height has a maximum. Fencing can be shaped into a long thin pen or a squarer one, so the enclosed area has a maximum. Lowering a price sells more items but earns less on each, so revenue has a maximum somewhere in between. In every case the strategy is the same: build an equation that models the situation, then find the vertex, since the vertex is exactly the maximum or minimum value being asked for. The other half of applied work is solving: "when does the ball land?" or "at what sales level do we break even?" are questions about where the model equals a particular value, and those use the methods from Solving Quadratic Equations.
Vertex question or solving question?
Most mistakes in applications come from doing the right algebra for the wrong question, so read the wording first. Words like maximum, minimum, greatest, least, optimal or best point to the vertex. The - (or -) coordinate of the vertex answers "when" or "at what input", and the -coordinate answers "how high", "how much" or "what is the largest value". Questions like "when does it hit the ground", "when is it 20 m high" or "at what sales level is profit zero" set the model equal to a value and solve. A parabola reaches most heights twice, once rising and once falling, so a solving question usually has two roots, and the context decides which ones to keep.
| The question asks | What you compute | Typical method |
|---|---|---|
| maximum height, largest area, greatest revenue | the y-coordinate of the vertex | , then substitute; or complete the square |
| when the maximum occurs, which price is best | the x-coordinate of the vertex | or the midpoint of the zeros |
| when it lands, when revenue is zero | the zeros of the model | factor or use the quadratic formula |
| when it reaches a given height, break-even points | solve model value | rearrange to = 0, then solve |
Projectile motion
Near Earth's surface, the height of an object under gravity alone follows , where is time in seconds, is the initial (launch) velocity in m/s, and is the initial height in metres. The comes from gravity ( with ) and is always negative, since gravity pulls the object back down. Each part of the model has a meaning you can read directly: is the launch height, the vertex gives the peak, and the positive zero is the landing time. The path is symmetric about the vertex, so an object launched from the ground spends as long rising as falling. When the zeros are rarely nice, so use the quadratic formula and reject the negative root, which describes a time before the launch. To find how long an object stays above a height, solve that height: for , setting gives , so and , and the ball is above m for seconds.
Area problems
When a fixed total (like a length of fencing) is split between two connected dimensions, expressing one dimension in terms of the other turns the area formula into a quadratic in a single variable — which can then be maximized. With m of fencing around a full rectangle, the two sides add to , so width gives length and area . The zeros are and (a pen with no width has no area), so the vertex is at the midpoint, : a square. Against a barn wall only three sides are fenced, so m gives length and , with zeros and and the best width . Always sketch and label the diagram first; most errors come from counting the wrong number of sides.
Revenue and profit problems
Revenue is always When price changes affect how many units sell (price goes down, sales go up), both price and quantity become linear expressions in the same variable, and their product is a quadratic to maximize. A handy shortcut avoids expanding: in , the zeros are and , so the vertex is at their midpoint . Profit is revenue minus cost. When revenue is quadratic and cost is linear, the break-even points are where the parabola meets the line. For and (in hundreds of dollars, for hundred items), setting gives , so or . The business makes a profit between those two sales levels, and the discriminant tells you whether break-even is even possible.