Interpreting a quadratic function model

A quadratic function can model a quantity that rises and then falls, or falls and then rises, such as the height of a thrown ball. The input usually measures time or distance, so the model applies only to a stated interval (its domain).

Each feature of the graph answers one question. The intercept on the vertical axis is the value when the input is \(0\): the starting value. The turning point gives the greatest or least value and when it happens, because a completed square \(-(t-p)^2+q\) can never be more than \(q\). Solving \(h=k\) finds when the quantity equals \(k\).

Method: put \(t=0\) for the starting value; complete the square to find the turning point; solve \(h=k\) for a given value (and \(h=0\) for when the quantity reaches zero); reject any value outside the domain, such as a negative time.

Example

The height, \(h\) metres, of a ball \(t\) seconds after it is thrown is modelled by \(h=8+2t-t^2\), from the throw until the ball lands.

\(\begin{aligned}h&=-(t^2-2t)+8\\&=-\left[(t-1)^2-1\right]+8\\&=-(t-1)^2+9\end{aligned}\)

Starting height: at \(t=0\), \(h=8\), so the ball is thrown from \(8\text{ m}\). Greatest height: \(-(t-1)^2\le0\), so \(h\le9\); the greatest height is \(9\text{ m}\), \(1\) second after the throw.

Landing: \(h=0\) gives \((t-1)^2=9\), so \(t=-2\) or \(t=4\). Reject \(t=-2\), which is before the throw; the ball lands after \(4\) seconds, so the model applies for \(0\le t\le4\).

Height \(5\text{ m}\): \((t-1)^2=4\) gives \(t=-1\) or \(t=3\). Only \(t=3\) lies in the domain, so the ball is \(5\text{ m}\) high once, on the way down; it started above \(5\text{ m}\).

The same features can be read from a plotted graph: the intercept on the vertical axis, the highest or lowest point, and the points where the horizontal line \(h=k\) meets the curve. Readings are estimates to the accuracy of the grid; the algebra above gives exact values.

Exam check: answer in the context, with units, and say why any root is rejected. A turning point outside the domain is not the greatest value in the situation.