Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 12.2
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Exercise 117 Page 688

Practice makes perfect
a We are given two intercepts, ( -3,0) and ( 2,0), that we want to use to write a quadratic function in standard form. Because the given points give us the roots of the function, we can begin by writing the equation in factored form. Then we can multiply the factors together.
Factored Form:& y=a(x-p)(x-q) Standard Form:& y=ax^2+bx+c In the factored form p and q are the roots of the function. Since we know that the roots are -3 and 2, we can partially write the factored form of our equation. y=a(x-( -3))(x- 2) ⇕ y=a( x+ 3)( x-2 ) Since a does not have any effect on the roots, we can choose any value other than 0. For simplicity we will let a= 1. y= 1(x+3)(x-2) ⇕ y=(x+3)(x-2) Finally, let's use the Distributive Property to obtain the standard form.
y=(x+3)(x-2)
â–Ľ
Multiply parentheses
y=x(x-2)+3(x-2)
y=x^2-2x+3(x-2)
y=x^2-2x+3x-6
y=x^2+x-6
Please note that this is just one example of a quadratic function that satisfies the given requirements.
b Similarly as in Part A we are given two intercepts, ( -3,0) and ( 12,0). Therefore, -3 and 12 will be roots of a quadratic function. This allows us to write an equation in factored form, and then we will change it to standard form by multiplying the factors.
y=a(x-( -3)) ( x- 12 ) ⇕ y=a(x+3)( x-1/2)We can choose any value of a except 0. Since one of the factors contains a fraction let a= 2, which is the denominator of the fraction. Then, our equation will have integer coefficients. y= 2( x+ 3)( x-1/2) Finally, let's use the Distributive Property to obtain the standard form.
y=2(x+3)(x-1/2)
y=2(x-1/2)(x+3)
y=(2x-1)(x+3)
â–Ľ
Multiply parentheses
y=(2x-1)x+(2x-1)3
y=2x^2-x+(2x-1)3
y=2x^2-x+6x-3
y=2x^2+5x-3
Once again, please note that this is just one example of a quadratic function that satisfies the given requirements.