Quadratic Models

Quadratic Models

Secondary 3

A quadratic model describes a quantity that rises to a greatest value and then falls, or falls to a least value and then rises: a height, an area, a revenue or the curve of an arch. Write the model in completed-square form \(y=a(x-h)^2+k\); its features answer most questions.

FeatureMeaning in the context
Turning point \((h,\,k)\)Greatest value \(k\) when \(a<0\), least value \(k\) when \(a>0\), reached when the input is \(h\). This value \(k\) is also called the stationary value: the value of the function at its turning point.
Value when the input is \(0\)Starting value, for example the height of a nozzle or a fixed cost
Roots of \(y=0\)When the height is zero (lands), or where an arch meets the ground
Line of symmetry \(x=h\)The highest (or lowest) point is halfway between any two points at the same height, such as the two feet of an arch or the two roots

Why: a fountain's water jet leaves a nozzle \(2\) m above the surface of a pool. Its height is \(y\) m above the water at a horizontal distance of \(x\) m from the nozzle, where \(y=2+12x-4x^2\), a downward parabola. Completing the square, \(y=-4(x-1.5)^2+11\): the squared term is never negative, so \(y\le11\), with equality only when \(x=1.5\). The model only applies from the nozzle, \(x=0\), to where the jet meets the water.

Method

  1. Read the model, or form it: name each variable with its unit, write one equation from the given fact (a total length, a perimeter, a price rule) and substitute.
  2. State the values the context allows, for example lengths greater than \(0\) or times from release until landing.
  3. Complete the square to find the greatest or least value and the input at which it occurs.
  4. To find when the model takes a given value, solve the equation and reject any solution outside the allowed values, saying why.
  5. Answer in words with units; give non-exact answers to 3 significant figures.

Example

A shop sells caps at \(\$x\) each and expects to sell \(60-2x\) caps a day. Its daily revenue is \(R=x(60-2x)=-2x^2+60x=-2\left[(x-15)^2-225\right]=-2(x-15)^2+450\). The greatest revenue is \(\$450\), at a price of \(\$15\).

For the jet above, it meets the water when \(y=0\): \((x-1.5)^2=\dfrac{11}{4}\), so \(x=1.5\pm\dfrac{\sqrt{11}}{2}\). Reject \(x=-0.158\), which is behind the nozzle at \(x=0\). The jet meets the water \(3.16\) m from the nozzle (3 s.f.). Its highest point, at \(x=1.5\), is not halfway to \(3.16\) m, because the jet starts \(2\) m above the water.

An "explain" or "justify" answer gives the deciding value and compares it with the requirement in one sentence: "The greatest height is \(11\) m, which is less than \(12\) m, so the jet cannot reach a sign \(12\) m above the water."

Exam check Every root of the equation must be checked against the context: a negative time or length is rejected with a reason, not left out silently. State units in the final answer.

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