Chapter 9.1, Problem 3BCYP

To determine

To Find:

The vertex , axis of symmetry and y − intercept.

Answer to Problem 3BCYP

Vertex : $(-\frac{1}{2},\frac{3}{2})$

Axis of Symmetry : $x=-\frac{1}{2}$

y-intercept : 2

Explanation of Solution

Given Information:

Given equation is :

  $y=2x^2+2x+2$

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

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

  The equation of axis of symmetry is $x=-\frac{1}{2}$

  • Find the Vertex :

  Since axis of symmetry passes through the vertex , we have x coordinate of the vertex (x,y) is $x=-\frac{1}{2}$

  Substituting the value of $x=-\frac{1}{2}$in the equation $y=2x^2+2x+2$

    $\begin{array}{l}y=2x^2+2x+2\\ $ y=2{(-\frac{1}{2})}^2+2(-\frac{1}{2})+\mathrm{2................}[x=-\frac{1}{2}]$y=\frac{1}{2}-1+\mathrm{2..................................}[simplify]\\ y=\frac{3}{2}\end{array}$

  The vertex is at $(-\frac{1}{2},\frac{3}{2})$

  • 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