Chapter 9, Problem 15MCQ

To determine

To graph and describe the relation between the given function to the graph of the function $h=t^2$ .

Explanation of Solution

Given information :

Given functions is $h=-16t^2+14$ .

Here h is the height of the box of nails after t seconds from when it is dropped.

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.

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 $h=-16t^2+14$ , the value of $\left|a\right|>1$ at $a=-16$ . Thus, the relation of $h=t^2$ when multiplied by a , as seen in the graph, is that it is stretched vertically by a factor of $16$ .

The graph represented by $g(x)=x^2+k$ 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 $h=-16t^2+14$ , the value of $k>0$ at $k=14$ . Therefore, the translation of the function with respect to the graph of $h=t^2$ , as seen in the graph shown above, can be described as a upward translation by $14$ units.

Therefore, in relation to the graph of $h=t^2$ , the graph of the function $h=-16t^2+14$ has been reflected along the x- axis, translated upwards by 14 units and stretched vertically by a factor of 16.