Chapter 4.7, Problem 56PPSE

Solution

a.

To Find: The year in which more than 100 million tons of carbon monoxide were released into the air.

The year in which more than 100 million tons of carbon monoxide were released into the air $1986$ and $1998$ .

Given:

The function $y=0.4409x^2-5.1724x+99.0321$

Initial year $=1987$

  $y=$ represents the amount of gas released

  $x=0$ represents the year $1987$ .

Explanation:

Find the year in which more than 100 million tons of carbon monoxide were released into the air.

Sketch the function $y=0.4409x^2-5.1724x+99.0321$ (Parabola)

To find the year in which more than 100 million tons of gas released, sketch $y=100$ (straight line parallel to $x-$ axis)

Graphical representation:

  

The $x-$ coordinates of the point of intersection between the two functions $y=0.4409x^2-5.1724x+99.0321$ and $y=100$ can be calculated by draw a perpendicular line from that point on $x-$ axis.

Parabolas meet straight lines at two points, first at $x=0$ and then at $x=11.92$

So, from $1986$ to $1998$ , there was a release of $100$ million tons of carbon monoxide

As shown in the graph, the parabola has a greater than $100$ $y-$ value for all $x$ values above $11.92$ .

According to this calculation, more than $100$ million tons of carbon monoxide were released into the air after the year $1987+12=1999$ .

So, year in which more than 100 million tons of carbon monoxide were released into the air $1986$ and $1998$ .

b.

To Find: The year in which more than 100 million tons of carbon monoxide were released into the air by using the quadratic formula.

The year in which more than 100 million tons of carbon monoxide were released into the air $1986$ and $1998$

Given:

The function $y=0.4409x^2-5.1724x+99.0321$

Initial year $=1987$

  $y=$ represents the amount of gas released

  $x=0$ represents the year $1987$ .

Explanation:

Calculate the value of $x$ for which $y=100$ :

  $\begin{array}{c}0.4409x^2-5.1724x+99.0321=100\\ 0.4409x^2-5.1724x+99.0321-100=0\end{array}$

Use the quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute the values $a=0.4409,b=-5.1724,c=-0.9679$ into the formula:

  $\begin{array}{l}x=\frac{-\left(-5.1724\right)\pm \sqrt{{\left(-5.1724\right)}^2-4\left(0.4409\right)\left(-0.9679\right)}}{2\left(0.4409\right)}\\ x=\frac{5.1724\pm \sqrt{28.46}}{0.8818}\\ x\approx -0.18,12\end{array}$

Eliminate the negative value.

Add $12$ with the initial year $1986$ :

  $1986+12=1998$

So, year in which more than 100 million tons of carbon monoxide were released into the air $1986$ and $1998$

c.

To Find: The method that preferred.

Graphing method was more preferred.

Given:

The function $y=0.4409x^2-5.1724x+99.0321$

Initial year $=1987$

  $y=$ represents the amount of gas released

  $x=0$ represents the year $1987$ .

Explanation:

Weighing both methods, the graphing method takes more time and effort.

If to choose between two method, graphing method should be used since it gives more accurate data.

Otherwise, choose the quadratic formula method.