All courses Math II · A-SSE.1.a 8 of 73
Interpret terms, factors, and coefficients in quadratic and exponential expressions.

Interpret the leading coefficient of a quadratic expression

Problem
h(t)=−16t²+48t+5 models height h in feet at time t seconds. Enter leading coefficient with units, sign/opening effect, quadratic scale or second derivative, contextual consequence, and a guard against interpreting it as an ordinary constant slope.
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Hint

Identify input/output units, then choose units for a so a·input² has output units.

In ax²+bx+c, sign(a) sets opening and f''=2a measures constant curvature. The term ax² must have the output's units.

Solution walkthrough

01

Identify the leading coefficient

\[\text{Leading}~\text{coefficient}~=~-16\]

The leading coefficient is the number multiplying the squared term \(t^2\).

02

Interpret the negative sign

\[-16<0\]

A negative leading coefficient means the parabola opens downward, so the height rises and then falls.

03

Connect it to the projectile context

\[h(t)~=~-16t^2+48t+5\]

In a height model measured in feet over seconds, the \(-16\) reflects gravity's downward effect.

04

State the interpretation

\[\text{leading}~\text{coefficient}=-16~\text{feet}/\text{second}²;~\text{sign}~\text{effect}=\text{parabola}~\text{opens}~\text{downward};~\text{curvature}=h''(t)=-32~\text{feet}/\text{second}²;~\text{physical}~\text{meaning}=-16~\text{is}~\text{one}-\text{half}~\text{the}~\text{constant}~\text{vertical}~\text{acceleration},~\text{not}~\text{the}~\text{acceleration}~\text{itself};~\text{consequence}=\text{height}~\text{has}~a~\text{maximum}\]

This explains both the shape of the graph and the meaning of the coefficient in context.

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Another way

  1. Check the graph shape mentally: a projectile height model should rise, reach a maximum, and then come back down.

!

Common mistake

Interpreting \(-16\) as the initial height instead of noticing that it is attached to \(t^2\) and controls the downward curvature.