Chapter 9, Problem 15SGR

a.

To determine

To evaluate whether the function has a maximum or minimum value.

Answer to Problem 15SGR

The function has a minimum value.

Explanation of Solution

Given information :

The function provided is $h=-16t^2+32t$

Here, h is the height of the rocket t seconds after it is launched.

:

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 $h=-16t^2+32t$ , the value of a is negative and thus it opens downwards. Thus, it has a maximum value.

b.

To determine

To state the maximum or minimum value

Answer to Problem 15SGR

The maximum value is 0.

Explanation of Solution

Given information :

The function provided is $h=-16t^2+32t$

Here, h is the height of the rocket t seconds after it is launched.

Formula used :

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:

1. Highest point (if $a<0$) and be called the maximum. Here, the graph opens downwards.
2. Lowest point (if $a>0$) and be called the minimum. Here, the graph opens upwards.

Here, a is less than zero, hence the parabola opens downwards, and has a maximum value.

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

Axis of symmetry for the given function is

  $x=-\frac{b}{2\times a}$

Calculation :

Substituting the values from the given equation.

  $t=-\frac{(32)}{2\times (-16)}$

Simplifying.

  $t=1$

This corresponds to the x -coordinate of the vertex

To graph a quadratic equation using table of values, compute the ‘h’ values for the corresponding ‘t’ values. For this, substitute the values of ‘t’ with negative as well as positive integers and check for the vertex.

For example, to compute the value of ‘h’ when $t=1$ , substitute the value of ‘t’ in equation $h=-16t^2+32t$

On substituting the value of the ‘t’ as $t=1$ , it is,

  $h=-16{(1)}^2+32(1)$

On simplifying this,

Finally, the value of ‘h’ is

  $=16$

The maximum value of the function is $h=16$ at $(1,16)$ .

c.

To determine

To state domain and range of the function

Answer to Problem 15SGR

The domain of this equation is $\left\{0<x<2\right\}$ and the range is $\left\{h|h\le 16\right\}$

Explanation of Solution

Given information :

The function provided is $h=-16t^2+32t$ .

Here, h is the height of the rocket t seconds after it is launched.

The axis of symmetry is the line $t=1$

The maximum value is $h=16$ .

The reasonable range would be the two x- intercepts if x- axis is considered to be the ground. Therefore, considering the values of t from the time the rocket is launched to the time it hits the ground, that is, where $h=0$ .

Substituting the value $h=0$ in the given function.

  $-16t^2+32t=0$

Solving in quadratic form.

  $t=\frac{-32\pm \sqrt{{(32)}^2-4\times (-16)\times 0}}{2(-16)}$

Simplifying.

  $t=\frac{-32\pm \sqrt{{(32)}^2}}{-32}$

Removing the root.

  $t=\frac{-32\pm 32}{-32}$

Thus, the value of $t=0,2$ .

Therefore, the domain of this equation is $\left\{0<x<2\right\}$ and the range is $\left\{y|y\le 16\right\}$ . This is because the vertex of the parabola is maximum and at 16.