Chapter 9.1, Problem 11CYU

To determine

To Find:

The vertex , axis of symmetry and y − intercept.

Answer to Problem 11CYU

Vertex : (2,1)

Axis of Symmetry : x = 2

y-intercept : 5

Explanation of Solution

Given Information:

Given equation is :

  $y=x^2-4x+5$

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 $y=x^2-4x+5$ ; comparing it with standard form of quadratic equation , we have a = 1 , b = -4 , c = 5 .
• Find Axis of Symmetry :

  $\begin{array}{l}y=a{x}^2+bx+c\mathrm{...........................}[s\tan dard\_form]\\ x=-\frac{b}{2a}\mathrm{.....................................}[formula\_of\_axis\_of\_symmetry]\\ x=-\frac{-4}{2(1)}=\mathrm{2..........................}[a=1,b=-4]\end{array}$

The equation of axis of symmetry is x = 2

• Find the Vertex :

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 $y=x^2-4x+5$

  $\begin{array}{l}y=x^2-4x+5\\ $ y={(2)}^2-4(2)+\mathrm{5.....................}[x=2]$y=4-8+\mathrm{5................................}[simplify]\\ y=1\end{array}$

The vertex is at (2,1)

• Find y-intercept :

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

Hence , substituting x=0 in the equation $y=x^2-4x+5$ , we will get the y-intercept.

  $\begin{array}{l}y=x^2-4x+5\\ $ y={(0)}^2-4(0)+\mathrm{5.....................}[x=0]$y=5\end{array}$

Or

y-intercept always occurs at (0,c) and c = 5 here.

So, y-intercept = 5 and is located at (0,5)