Chapter 9, Problem 11SGR

a.

To determine

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

Answer to Problem 11SGR

The function has a minimum value.

Explanation of Solution

Given information :

The function provided is $y=x^2-4x+4$

:

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 greater than 0 hence the graph will have a minimum and will open upwards.

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

In the function $y=x^2-4x+4$ , the value of a is positive and thus it opens upwards. Thus, it has a minimum value.

b.

To determine

To state the maximum or minimum value

Answer to Problem 11SGR

The minimum value is 0.

Explanation of Solution

Given information :

The function provided is $y=x^2-4x+4$

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 greater than zero, hence the parabola opens upwards, and has a minimum 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.

  $x=-\frac{(-4)}{2\times (1)}$

Simplifying.

  $x=2$

This corresponds to the x -coordinate of the vertex

To graph a quadratic equation using table of values, compute the ‘y’ values for the corresponding ‘x’ values. For this, substitute the values of ‘x’ with negative as well as positive integers and check for the vertex.

For example, to compute the value of ‘y’ when $x=2$ , substitute the value of ‘x’ in equation $y=x^2-4x+4$

On substituting the value of the ‘x’ as $x=2$ , it is,

  $y={(2)}^2-4\times (2)+4$

On simplifying this,

  $=4-8+4$

Finally, the value of ‘y’ is

  $=0$

The minimum value of the function is $y=0$ at $(2,0)$ .

c.

To determine

To state domain and range of the function

Answer to Problem 11SGR

The domain of this equation is $\left\{-\infty <x<\infty \right\}$ and the range is $\left\{y|y\ge 0\right\}$

Explanation of Solution

Given information :

The function provided is $y=x^2-4x+4$ .

The axis of symmetry is the line $x=2$

The minimum value is $y=0$ .

Since the parabola extends to infinity, the domain of this equation is $\left\{-\infty <x<\infty \right\}$ and the range is $\left\{y|y\ge 0\right\}$ . This is because the vertex of the parabola is minimum and at 0.