Chapter 9.1, Problem 19CYU

To determine

To Graph:

The given function.

Explanation of Solution

Given Information:

Given equation is :

  $y=2x^2-8x-4$

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.

• 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.

Calculation:

• In the equation $y=2x^2-8x-4$ ; comparing it with standard form of quadratic equation , we have a = 2, b = -8 , c = -4 .
• Find equation of Axis of Symmetry :

  $\begin{array}{l}y=a{x}^2+bx+c\mathrm{...........................}[s\tan dard\_form]\\ x=-\frac{b}{2a}\mathrm{.....................................}[formula\_of\_axis\_of\_symmetry]\\ x=-\frac{-8}{2(2)}=\mathrm{2..........................}[a=2,b=-8]\end{array}$

The equation of axis of symmetry is $x=2$

• Find the Vertex and determine whether it is maximum or minimum:

Since axis of symmetry passes through the vertex , we have x coordinate of the vertex (x,y) is $x=2$

Substituting the value of $x=2$in the equation $y=2x^2-8x-4$

  $\begin{array}{l}y=2x^2-8x-4\\ $ y=2{(2)}^2-8(2)-\mathrm{4...............}[x=2]$y=8-16-\mathrm{4..................................}[simplify]\\ y=-12\end{array}$

The vertex is at $(2,-12)$ .

Since the a = 2 > 0 .So, the graph opens upward .Hence, the vertex is the minimum.

• Find y-intercept :

The y-coordinate of the point at which the graph cuts the y-axis is the y-intercept.

y-intercept always occurs at (0,c) and c = -4 here.

So, y-intercept = -4 and point is (0,-4)

• The axis of symmetry cuts the parabola into two equal parts . So, if there is a point on one side, there is a corresponding point on the other side that is at a same distance from the axis of symmetry and the same y-value.

x	y
0	-4
1	-10
2	-12
3	-10
4	-4

• Connect the points with a smooth curve.

Graph :