Chapter 9.1, Problem 50PPS

(a)

To determine

Whether the function has a maximum or a minimum value.

Answer to Problem 50PPS

The given function has a minimum value.

Explanation of Solution

Given Information:

Given equation is :

  $y=2x^2+4x+6$

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• If a > 0 ,
• The graph of $a{x}^2+bx+c$ opens upward.
• The lowest point on the graph is the minimum.
• If a < 0 ,
• The graph of $a{x}^2+bx+c$ opens downward.
• The highest point on the graph is the maximum.
• The maximum or the minimum is the vertex.

Calculation:

• In the equation $y=2x^2+4x+6$ ; comparing it with standard form of quadratic equation , we have a = 2 , b = 4 , c = 6 .
• Since the value of a is 2 , which is greater than 0. So, the graph opens upward and hence the function has minimum value.

(b)

To determine

To Find:

The maximum or minimum value of the function.

Answer to Problem 50PPS

The minimum value is 4.

Explanation of Solution

Given Information:

Given equation is :

  $y=2x^2+4x+6$

And from part (a) , we have that the function has minimum value.

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• Shape of the graph of a Quadratic function is called Parabola.
• Parabolas are symmetric about a central line called Axis Of Symmetry.
• The axis of symmetry intersects a parabola only at one point , called the Vertex.
• Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves.

  $x=-\frac{b}{2a}$

Calculation:

• In the equation $y=2x^2+4x+6$; comparing it with standard form of quadratic equation , we have a = 2 , b = 4 , c = 6 .
• To find the minimum value:
• The minimum value is the y-coordinate of the vertex .
• The x-coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{4}{2(2)}=-1$
• Substitute x = -1 in the given function to find the y-coordinate of the vertex :

  $\begin{array}{l}y=2x^2+4x+6\\ $ y=2{(-1)}^2+4(-1)+\mathrm{6.........................}[x=-1]$y=2-4+6\\ y=-2+6=4\end{array}$

So, the vertex is (-1,4)

So, the minimum value of the function is 4.

(c)

To determine

To Find:

The domain and range of the function.

Answer to Problem 50PPS

Domain is all real numbers.

Range = $\{y\in R:y\ge 4\}$

Where R= set of all real numbers.

Explanation of Solution

Given Information:

Given equation is :

  $y=2x^2+4x+6$

From part (b) , the vertex of the given function is (-1,4) and minimum of the function is at 4.

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $f(x)=a{x}^2+bx+c;a\ne 0$

This is the standard form of Quadratic function.

• The range of a quadratic function is all real numbers greater than or equal to the minimum , or all real numbers less than or equal to the maximum.

Calculation:

• In the equation $y=2x^2+4x+6$ ; comparing it with standard form of quadratic equation , we have a = 2 , b = 4, c = 6 .
• The vertex is (-1,4).
• The domain of the function is all real numbers .
• The range of a quadratic function is all real numbers greater than or equal to the minimum.So, the range is $\{y\in R:y\ge 4\}$ where R= set of all real numbers.