Finding Unknown Inputs

Student Summary

The height of a softball, in feet, tt seconds after someone throws it straight up, can be defined by f(t)=-16t2+32t+5f(t) = \text-16t^2+32t+5. The input of function ff is time, and the output is height.

We can find the output of this function at any given input. For instance:

  • At the beginning of the softball's journey, when t=0t = 0, its height is given by f(0)f(0).
  • Two seconds later, when t=2t=2, its height is given by f(2)f(2).

The values of f(0)f(0) and f(2)f(2) can be found using a graph or by evaluating the expression -16t2+32t+5\text-16t^2+32t+5 at those values of tt. What if we know the output of the function and want to find the inputs? For example:

  • When does the softball hit the ground?

    Answering this question means finding the values of tt that make f(t)=0f(t)=0, or solving -16t2+32t+5=0\text -16t^2+32t+5=0

  • How long will it take the ball to reach 8 feet?

    This means finding one or more values of tt that make f(t)=8f(t) = 8, or solving the equation -16t2+32t+5=8\text -16t^2+32t+5=8.

The equations -16t2+32t+5=0\text -16t^2+32t+5=0 and -16t2+32t+5=8\text -16t^2+32t+5=8 are quadratic equations. One way to solve these equations is by graphing y=f(t)y = f(t).

  • To answer the first question, we can look for the horizontal intercepts of the graph, where the vertical coordinate is 0.
  • To answer the second question, we can look for the horizontal coordinates that correspond to a vertical coordinate of 8.

<p>Parabola facing down. X intercept at 6. Maximum at about 2 point 8 comma 150. points plotted at 0 point 1 comma 8, 1 point 9 comma 8, and 2 point 1 5 comma 0. </p>

We can see that there are two solutions to the equation -16t2+32t+5=8\text -16t^2+32t+5=8 and one solution to the equation -16t2+32t+5=0\text{-}16t^2+32t+5=0.

The softball has a height of 8 feet twice, when going up and when coming down, and these instances occur when tt is about 0.1 and 1.9. It has a height of 0 once, when tt is about 2.15.

Often, when we are modeling a situation mathematically, an approximate solution is good enough. Sometimes, however, we would like to know exact solutions, and it may not be possible to find them using a graph. In this unit, we will learn more about quadratic equations and how to solve for exact answers using algebraic techniques.

Visual / Anchor Chart

Standards

Building On
F-IF.4

F-IF.4

F-IF.4

F-IF.4

F-IF.4

F-IF.4

F-LE.6

F-LE.6

HSF-IF.B.4

Addressing
A-CED.1

A-CED.1

A-CED.1

A-CED.1

A-CED.1

A-CED.1

HSA-CED.A.1

Building Toward
A-CED.3

A-CED.3

A-CED.3

A-CED.3

A-CED.3

HSA-CED.A.3