McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
1. Solving Quadratic Equations by Factoring
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Exercise 68 Page 175

Practice makes perfect
a Let's begin by substituting a= 1, p=2, and q=-3 in the equation. Then we can simplify it.
f(x)=a(x-p)(x-1)
f(x)= 1(x-2)(x-(-3))
â–Ľ
Simplify
f(x)=1(x-2)(x+3)
f(x)=(x-2)(x+3)
f(x)=x(x-2)+3(x-2)
f(x)=x^2-2x+3(x-2)
f(x)=x^2-2x+3x-6
f(x)=x^2+x-6
Notice that the leading coefficient is positive. Therefore, the graph opens upward and it has a minimum value on its vertex. To graph the equation, we will first find its axis of symmetry by using the formula x=- b2 a. Let's first highlight the coefficient of the function and determine a and b. f(x)= 1x^2+ 1x -6 Therefore, a= 1 and b= 1. Now, we can substitute these values into the formula, so we can find the axis of symmetry.
x=-b/2a
x=-1/2( 1)
x=-1/2
Next, we will make a table of values using x values around the axis of symmetry which is x=- 12.
x x^2+x-6 f(x)
-4 ( -4)^2+( -4)-6 6
-3 ( -3)^2+( -3)-6 0
-2 ( -2)^2+( -2)-6 -4
-1 ( -1)^2+( -1)-6 -6
-1/2 ( -1/2)^2+( -1/2)-6 -25/4
0 ( 0)^2+ 0-6 -6
1 ( 1)^2+ 1-6 -4
2 ( 2)^2+ 2-6 0
3 ( 3)^2+ 3-6 6

Let's plot the points and draw a parabola that passes through these points and complete the graph of the function.

b When we substituted the given values, we obtained the factorized form of the equation in Part A.
(x-2)(x+3)=0 Let's solve it by the Zero Product Property to find its solutions.
(x-2)(x+3)=0
lx-2=0 x+3=0
lx=2 x+3=0
lx=2 x=-3
Therefore, the solutions of the equation are 2 and -3.
c We will begin by determining the functions for a=4, -3 and 12.
a a(x-2)(x+3) f(x)
4 4(x-2)(x+3) 4x^2+4x-24
-3 -3(x-2)(x+3) -3x^2-3x+18
1/2 1/2(x-2)(x+3) 1/2x^2+1/2x-3

To graph the functions, we will proceed in the same way as we did in Part A. Let's first find their axes of symmetry.

f(x) x=-b/2a Axis of Symmetry
4x^2+ 4x -24 x=-4/2( 4) x=-1/2
-3x^2 -3x+ 18 x=--3/2( -3) x=-1/2
1/2x^2+ 1/2x -3 x=-1/2/2( 1/2) x=-1/2

Now, we can graph these functions as we did in Part A. Notice that the leading coefficient of the second function is negative, so the graph of it opens downward. The other two functions open upward because of the positive leading coefficient.

d By looking at the graphs in Part C, we can list the similarities and differences as shown below.
  • All three curves have the same zeros and the same axes of symmetry.
  • The graphs with a positive leading coefficients open upward and the graph with negative leading coefficient opens downward.
  • The graphs with leading coefficients a=4 and a=-3 are much steeper than the graph with a leading coefficient a= 12.
e Let's consider the factored form of a quadratic function.

f(x)=a(x-p)(x-q) When a quadratic function is in factored form as shown, we can immediately determine its solutions as p and q by the Zero Product Property.