Chapter 9.2, Problem 19PPS

a.

To determine

To describe and graph the translation of the function $c(x)$ and $p(x)$ in relation to the graph of parent function $f(x)=x^2$ .

Explanation of Solution

Given information :

Given functions are $c(x)=-16{(x-2.5)}^2+105$ and $p(x)=-16{(x-2.8)}^2+126.5$ .

Here $c(x)$ and $p(x)$ are the height of the rocket of “C” and “P” respectively, after x seconds.

Graph :

  

Interpretation :

The graph of quadratic function representing the cross section of the football field is compared to the graph of $f(x)=x^2$ .

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. The parabola is symmetric in nature and the point of symmetry is known as the vertex. This vertex lies on the line of symmetry and is symmetric about this line.

This vertex point shall be:

Highest point (if $a<0$ ) and be called the maximum. Here, the graph opens downwards.

Or, lowest point (if $a>0$ ) and can be called the minimum. Here the graph opens upwards.

In this case, ‘a’ is lesser than 0 hence the graph will have a maximum and will open downwards.

A parabola always points to infinity, either negative or positive.

In the function $c(x)=-16{(x-2.5)}^2+105$ , the value of a is negative and thus it opens downwards. Therefore, the transformation of the function with respect to the graph of $f(x)=x^2$ , as seen in the graph shown above, is that the function has been inverted and now it open downwards.

The graph of quadratic function $f(x)$ when multiplied by a positive constant a , the resulting graph $af(x)$ is a vertical dilation of $f(x)$ . The function is stretched or compressed vertically by a factor of $\left|a\right|$ .

If $\left|a\right|$ $>1$ , the graph of $f(x)$ is stretched vertically, that is, all points on the graph $f(x)$ move farther away from the x -axis.

If $\left|a\right|<1$ , the graph of $f(x)$ is compressed vertically, that is, all points on the graph $f(x)$ move closer to the x -axis.

In the function $c(x)=-16{(x-2.5)}^2+105$ , the value of $\left|a\right|>1$ at $a=-16$ . Thus, the transformation of $f(x)$ when multiplied by a , as seen in the graph, is that it is stretched vertically by a factor of $16$ .

The graph represented by $g(x)={(x-h)}^2$ is equivalent to the graph of $f(x)=x^2$ that has been translated horizontally. The direction of translation or movement of the graph is dependent on the value of $h$ .

If $h>0$ , the graph of $f(x)=x^2$ is translated $\left|h\right|$ units to the right.

If $h<0$ , the graph of $f(x)=x^2$ is translated $\left|h\right|$ units to the left.

In the function $c(x)=-16{(x-2.5)}^2+105$ , the value of $h>0$ at $h=2.5$ . Therefore, the translation of the function with respect to the graph of $f(x)=x^2$ , as seen in the graph shown above, is $2.5$ units to the right.

The graph represented by $g(x)=x^2+k$ is equivalent to the graph of $f(x)=x^2$ that has been translated vertically. The direction of translation or movement of the graph is dependent on the value of k .

If $k>0$ , the graph of $f(x)=x^2$ is translated $\left|k\right|$ units upwards.

If $k<0$ , the graph of $f(x)=x^2$ is translated $\left|k\right|$ units downwards.

In the function $c(x)=-16{(x-2.5)}^2+105$ , the value of $k>0$ at $k=105$ . Therefore, the translation of the function with respect to the graph of $f(x)=x^2$ , as seen in the graph shown above, can be described as a upward translation by $105$ units.

Therefore, as seen in the graph, the transformation of the function $c(x)=-16{(x-2.5)}^2+105$ with respect to the graph of $f(x)=x^2$ is the inversion of the graph, horizontal translation $2.5$ units to the right, vertical stretch by a factor of $16$ and a vertical translation upwards by $105$ units.

The graph of quadratic function representing the cross section of the football field is compared to the graph of $f(x)=x^2$ .

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. The parabola is symmetric in nature and the point of symmetry is known as the vertex. This vertex lies on the line of symmetry and is symmetric about this line.

This vertex point shall be:

Highest point (if $a<0$ ) and be called the maximum. Here, the graph opens downwards.

Or, lowest point (if $a>0$ ) and can be called the minimum. Here the graph opens upwards.

In this case, ‘a’ is lesser than 0 hence the graph will have a maximum and will open downwards.

A parabola always points to infinity, either negative or positive.

In the function $p(x)=-16{(x-2.8)}^2+126.5$ , the value of a is negative and thus it opens downwards. Therefore, the transformation of the function with respect to the graph of $f(x)=x^2$ , as seen in the graph shown above, is that the function has been inverted and now it opens downwards.

The graph of quadratic function $f(x)$ when multiplied by a positive constant a , the resulting graph $af(x)$ is a vertical dilation of $f(x)$ . The function is stretched or compressed vertically by a factor of $\left|a\right|$ .

If $\left|a\right|$ $>1$ , the graph of $f(x)$ is stretched vertically, that is, all points on the graph $f(x)$ move farther away from the x -axis.

If $\left|a\right|<1$ , the graph of $f(x)$ is compressed vertically, that is, all points on the graph $f(x)$ move closer to the x -axis.

In the function $p(x)=-16{(x-2.8)}^2+126.5$ , the value of $\left|a\right|>1$ at $a=-16$ . Thus, the transformation of $f(x)$ when multiplied by a , as seen in the graph, is that it is stretched vertically by a factor of $16$ .

The graph represented by $g(x)={(x-h)}^2$ is equivalent to the graph of $f(x)=x^2$ that has been translated horizontally. The direction of translation or movement of the graph is dependent on the value of $h$ .

If $h>0$ , the graph of $f(x)=x^2$ is translated $\left|h\right|$ units to the right.

If $h<0$ , the graph of $f(x)=x^2$ is translated $\left|h\right|$ units to the left.

In the function $p(x)=-16{(x-2.8)}^2+126.5$ , the value of $h>0$ at $h=2.8$ . Therefore, the translation of the function with respect to the graph of $f(x)=x^2$ , as seen in the graph shown above, is $2.8$ units to the right.

The graph represented by $g(x)=x^2+k$ is equivalent to the graph of $f(x)=x^2$ that has been translated vertically. The direction of translation or movement of the graph is dependent on the value of k .

If $k>0$ , the graph of $f(x)=x^2$ is translated $\left|k\right|$ units upwards.

If $k<0$ , the graph of $f(x)=x^2$ is translated $\left|k\right|$ units downwards.

In the function $p(x)=-16{(x-2.8)}^2+126.5$ , the value of $k>0$ at $k=126.5$ . Therefore, the translation of the function with respect to the graph of $f(x)=x^2$ , as seen in the graph shown above, can be described as a upward translation by $126.5$ units.

Therefore, as seen in the graph, the transformation of the function $p(x)=-16{(x-2.8)}^2+126.5$ with respect to the graph of $f(x)=x^2$ is the inversion of the graph, horizontal translation $2.8$ units to the right, vertical stretch by a factor of $16$ and a vertical translation upwards by $126.5$ units.

b.

To determine

To calculate which rocket went higher.

Answer to Problem 19PPS

Rocket of “P” went higher.

Explanation of Solution

Given information :

Given functions are $c(x)=-16{(x-2.5)}^2+105$ and $p(x)=-16{(x-2.8)}^2+126.5$ .

Here $c(x)$ and $p(x)$ are the height of the rocket of “C” and “P” respectively, after x seconds.

:

Here, vertex is $(h,k)$ . A quadratic equation can be defined as:

  $f(x)=a{(x-h)}^2+k$

Hence, it is evident that $p(x)$ has a higher k value at $k=126.5$ .

Since, k is the y- coordinate of the vertex, it concluded that $p(x)=-16{(x-2.8)}^2+126.5$ is higher.

c.

To determine

To state which rocket stayed in the air longer.

Answer to Problem 19PPS

Rocket of “P” stayed in the air longer.

Explanation of Solution

Given information :

Given functions are $c(x)=-16{(x-2.5)}^2+105$ and $p(x)=-16{(x-2.8)}^2+126.5$ .

Here $c(x)$ and $p(x)$ are the height of the rocket of “C” and “P” respectively, after x seconds.

Graph :

  

Interpretation :

As seen in the graph rocket of “C” hits the ground before rocket of “P”. Thus, the rocket of “P” stays in the air longer.