Chapter 9.1, Problem 74HP

To determine

To Explain:

How to find the axis of symmetry from an equation for a quadratic equation.What other characteristics of the graph can you derive from the equation.

Answer to Problem 74HP

We can find axis of symmetry , vertex and y-intercept of the graph from the given quadratic equation.

Explanation of Solution

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:

Let the quadratic equation be $y=-3x^2+6x-5$

Standard form of Quadratic equation is :

  $y=a{x}^2+bx+c$

In the equation $y=-3x^2+6x-5$; comparing it with standard form of quadratic equation , we have a = -3 , b = 6 , c = -5 .

Find Axis of Symmetry :

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

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

In the example taken , we find the axis of symmetry:

  $\begin{array}{l}y=a{x}^2+bx+c\mathrm{...........................}[s\tan dard\_form]\\ x=-\frac{b}{2a}\mathrm{.....................................}[formula\_of\_axis\_of\_symmetry]\\ x=-\frac{6}{2(-3)}=\mathrm{1..........................}[a=-3,b=6]\end{array}$

So, the axis of symmetry is x = 1

We can also find the vertex from the equation of a quadric equation.

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

  $x=-\frac{b}{2a}$ .

Substituting the value of $x=-\frac{b}{2a}$ in the equation and solving for y , we get the coordinates of the vertex.

Find the Vertex in the example :

Substituting the value of x =1 in the equation $y=-3x^2+6x-5$

  $\begin{array}{l}y=-3x^2+6x-5\\ $ y=-3{(1)}^2+6(1)-\mathrm{5.....................}[x=1]$y=-3+6-\mathrm{5................................}[simplify]\\ y=-2\end{array}$

The vertex is at (1,-2)

We can also find the y-intercept of the graph. Since 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 , we can find the y-intercept.

Find y-intercept in the example:

Substituting x=0 in the equation $y=-3x^2+6x-5$ , we will get the y-intercept.

  $\begin{array}{l}y=-3x^2+6x-5\\ $ y=-3{(0)}^2+6(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

Hence , we can find axis of symmetry , vertex and y-intercept of the graph from the given quadratic equation.