Chapter 9.1, Problem 4BGP

To determine

To Find:

The maximum or minimum value of the function.

Answer to Problem 4BGP

The minimum value is -3.

Explanation of Solution

Given Information:

Given equation is :

  $y=2x^2-4x-1$

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.
• Parent Function:

  $y=x^2$

• Standard Form :

  $f(x)=a{x}^2+bx+c;a\ne 0$

• Type of Graph:

Parabola

• Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves.

  $x=-\frac{b}{2a}$

• y-intercept = The y coordinate of the point at which the graph intersects y-axis = c
• Graph:

  

• 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.
• 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 $y=2x^2-4x-1$ ; comparing it with standard form of quadratic equation , we have a = 2 , b = -4 , c = -1 .
• Since the value of a is 2 , which is greater than 0. So, the graph opens upward and hence the function has minimum value.
• 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-1\\ $ y=2{(1)}^2-4(1)-\mathrm{1.........................}[x=1]$y=2-4-1\\ y=-3\end{array}$

So, the minimum value of the function is -3.