Chapter 9.1, Problem 39PPS

To determine

To Find:

The vertex , axis of symmetry and y − intercept.

Answer to Problem 39PPS

Vertex : (2,-14)

Axis of Symmetry : x = 2

y-intercept : 14

Explanation of Solution

Given Information:

Given equation is :

  $y=7x^2-28x+14$

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 $y=7x^2-28x+14$ ; comparing it with standard form of quadratic equation , we have a = 7 , b = -28, c = 14 .
• Find Axis of Symmetry :

  $\begin{array}{l}y=a{x}^2+bx+c\mathrm{...........................}[s\tan dard\_form]\\ x=-\frac{b}{2a}\mathrm{.....................................}[formula\_of\_axis\_of\_symmetry]\\ x=-\frac{-28}{2(7)}=\mathrm{2..........................}[a=7,b=-28]\end{array}$

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 $y=7x^2-28x+14$

  $\begin{array}{l}y=7x^2-28x+14\\ $ y=7{(2)}^2-28(2)+\mathrm{14.....................}[x=2]$y=28-28(2)+\mathrm{14................................}[simplify]\\ y=-28+14=-14\end{array}$

The vertex is at (2,-14)

• 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 $y=7x^2-28x+14$ , we will get the y-intercept.

  $\begin{array}{l}y=7x^2-28x+14\\ $ y=7{(0)}^2-28(0)+\mathrm{14.........................}[x=0]$y=14\end{array}$

Or

y-intercept always occurs at (0,c) and c = 14 here.

So, y-intercept = 14 and the point is (0,14).