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All courses Algebra I · A-SSE.1.a 54 of 76
Interpret terms, factors, and coefficients in quadratic and exponential expressions.

Interpret the leading coefficient of a quadratic expression

Problem
Analyze the leading coefficient in \(h(t)=-16t^{2}+48t+5\), with height in feet and time in seconds. State its units, graph direction, relation to the constant second derivative, physical meaning, and turning-point consequence.
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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 coefficient and its units

\[\begin{aligned} h(t)=-16t^2+48t+5 \\ a=-16~\text{feet}/\text{second}^2 \end{aligned}\]

The term a t squared must have height units of feet. Because t squared has units seconds squared, a has units feet per second squared.

02

Use the sign to read the graph

\[a=-16<0~->~\text{parabola}~\text{opens}~\text{downward}\]

A negative leading coefficient makes quadratic outputs decrease away from the vertex, so the height graph opens downward.

03

Relate the coefficient to acceleration

\[\begin{aligned} h'(t)=-32t+48 \\ h''(t)=-32~\text{feet}/\text{second}^2=2a \end{aligned}\]

Differentiating the quadratic twice shows the constant vertical acceleration. Therefore -16 is one-half of -32, not the full acceleration.

04

Interpret the turning point

\[\text{opens}~\text{downward}~->~\text{vertex}~\text{is}~a~\text{maximum}~\text{height}\]

A downward-opening parabola rises to its vertex and then falls, so the physical height has a maximum. This connects the coefficient's sign, curvature, acceleration meaning, and extremum.

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Another way

  1. Compare h(t) with h0+v0 t+(1/2)g t²; then (1/2)g=-16 gives g=-32 feet per second squared.

!

Common mistake

Do not call -16 the acceleration. In h(t)=at²+bt+c, the constant second derivative is 2a, here -32 feet per second squared.

Solution walkthrough video