Chapter 9.1, Problem 56PPS

To determine

To Graph:

The given function.

Explanation of Solution

Given Information:

Given equation is :

  $y=x^2-2x+2$

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 $y=x^2-2x+2$ ; comparing it with standard form of quadratic equation , we have a = 1 , b = -2 , c = 2 .
• Find equation of Axis of Symmetry :

  $\begin{array}{l}y=a{x}^2+bx+c\mathrm{...........................}[s\tan dard\_form]\\ x=-\frac{b}{2a}\mathrm{.....................................}[formula\_of\_axis\_of\_symmetry]\\ x=-\frac{-2}{2(1)}=\mathrm{1..........................}[a=1,b=-2]\end{array}$

The equation of axis of symmetry is $x=1$

• 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 $x=1$

Substituting the value of $x=1$in the equation $y=x^2-2x+2$

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

The vertex is at $(1,1)$ .

Since the a = 1 > 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 = 2 here.

So, y-intercept = 2 and it occurs at the point (0,2).

• 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
-1	5
0	2
1	1
2	2
3	5

• Connect the points with a smooth curve.

Graph :