Chapter 4.1, Problem 47PPSE

Solution

a.

To Find: The $y-$ value for each value of $x$, $\left(0,\boxed{}\right)\text{ and }\left(1,\boxed{}\right)$ .

The points are $\left(0,10\right)\text{ and }\left(\text{1,9}\right)$ .

The rate of change is $-1$ .

Given: The graph of the equation,

  

Explanation:

Draw a vertical line at $x=0$ and $x=1$ then check the $y-$ values on the graph.

  

From the graph, the $y-$ values are $10\text{ and 9}$ .

Thus, the point is $\left(0,10\right)\text{ and }\left(\text{1,9}\right)$ .

Now calculate the rate of change between the points.

  $\begin{array}{c}\text{rate}=\frac{9-10}{1-0}\\ =\frac{-1}{1}\\ =-1\end{array}$

b.

To Find: The $y-$ value for each value of $x$, $\left(1,\boxed{}\right)\text{ and }\left(2,\boxed{}\right)$ .

The points are $\left(2,6\right)\text{ and }\left(\text{1,9}\right)$ .

The rate of change is $-3$ .

Given: The graph of the equation,

  

Explanation:

Draw a vertical line at $x=2$ and $x=1$ then check the $y-$ values on the graph.

  

From the graph, the $y-$ values are $6\text{ and 9}$ .

Thus, the point is $\left(2,6\right)\text{ and }\left(\text{1,9}\right)$ .

Now calculate the rate of change between the points.

  $\begin{array}{c}\text{rate}=\frac{9-6}{1-2}\\ =\frac{3}{-1}\\ =-3\end{array}$

c.

To Find: The $y-$ value for each value of $x$, $\left(3,\boxed{}\right)\text{ and }\left(2,\boxed{}\right)$ .

The points are $\left(2,6\right)\text{ and }\left(3,1\right)$ .

The rate of change is $-5$ .

Given: The graph of the equation,

  

Explanation:

Draw a vertical line at $x=2$ and $x=3$ then check the $y-$ values on the graph.

  

From the graph, the $y-$ values are $6\text{ and 1}$ .

Thus, the point is $\left(2,6\right)\text{ and }\left(3,1\right)$ .

Now calculate the rate of change between the points.

  $\begin{array}{c}\text{rate}=\frac{1-6}{3-2}\\ =\frac{-5}{1}\\ =-5\end{array}$

d.

The change in rate of change away from the vertex.

Given: The graph of the equation,

  

Explanation:

The rate of change of three intervals are,

  $-1,-3,-5$

As the interval away from the vertex the rate is decreasing.

e.

The change in rate of change away from the vertex on the left side.

Given: The graph of the equation,

  

Explanation:

The rate of change of three intervals on right side of the vertex,

  $-1,-3,-5$

As the interval away from the vertex the rate is decreasing.

But if these points take to left side of the vertex. The rate is decreasing.