Question

quiz review 2- converting quadratic functions
instructions - given one form of the quadratic function, convert to the other two forms. you must show work on this paper
for credit.

	1	2	3	4	5	6
vertex form			$y=4(x - 3)^2 - 64$			$y=3(x - 5)^2 - 27$
factored form	$y=5(x + 2)(x - 6)$				$y=5(x - 9)(x + 5)$

Explanation:

Problem 1:

Step1: Expand factored to standard

First expand $(x+2)(x-6)$, then multiply by 5.
$y=5(x^2-6x+2x-12)=5(x^2-4x-12)=5x^2-20x-60$

Step2: Convert standard to vertex

Complete the square for $5x^2-20x-60$.
Factor 5 from first two terms: $y=5(x^2-4x)-60$
Complete the square: $x^2-4x=(x-2)^2-4$
Substitute back: $y=5[(x-2)^2-4]-60=5(x-2)^2-20-60=5(x-2)^2-80$

Problem 2:

Step1: Convert standard to vertex

Complete the square for $2x^2-8x-24$.
Factor 2 from first two terms: $y=2(x^2-4x)-24$
Complete the square: $x^2-4x=(x-2)^2-4$
Substitute back: $y=2[(x-2)^2-4]-24=2(x-2)^2-8-24=2(x-2)^2-32$

Step2: Convert standard to factored

Factor the standard form $2x^2-8x-24$.
First factor out 2: $y=2(x^2-4x-12)$
Factor quadratic: $x^2-4x-12=(x-6)(x+2)$
So $y=2(x-6)(x+2)$

Problem 3:

Step1: Expand vertex to standard

Expand $(x-3)^2$, multiply by 4, then subtract 64.
$y=4(x^2-6x+9)-64=4x^2-24x+36-64=4x^2-24x-28$

Step2: Convert standard to factored

Factor the standard form $4x^2-24x-28$.
First factor out 4: $y=4(x^2-6x-7)$
Factor quadratic: $x^2-6x-7=(x-7)(x+1)$
So $y=4(x-7)(x+1)$

Problem 4:

Step1: Convert standard to vertex

Complete the square for $5x^2-30x+40$.
Factor 5 from first two terms: $y=5(x^2-6x)+40$
Complete the square: $x^2-6x=(x-3)^2-9$
Substitute back: $y=5[(x-3)^2-9]+40=5(x-3)^2-45+40=5(x-3)^2-5$

Step2: Convert standard to factored

Factor the standard form $5x^2-30x+40$.
First factor out 5: $y=5(x^2-6x+8)$
Factor quadratic: $x^2-6x+8=(x-2)(x-4)$
So $y=5(x-2)(x-4)$

Problem 5:

Step1: Expand factored to standard

First expand $(x-9)(x+5)$, then multiply by 5.
$y=5(x^2+5x-9x-45)=5(x^2-4x-45)=5x^2-20x-225$

Step2: Convert standard to vertex

Complete the square for $5x^2-20x-225$.
Factor 5 from first two terms: $y=5(x^2-4x)-225$
Complete the square: $x^2-4x=(x-2)^2-4$
Substitute back: $y=5[(x-2)^2-4]-225=5(x-2)^2-20-225=5(x-2)^2-245$

Problem 6:

Step1: Expand vertex to standard

Expand $(x-5)^2$, multiply by 3, then subtract 27.
$y=3(x^2-10x+25)-27=3x^2-30x+75-27=3x^2-30x+48$

Step2: Convert standard to factored

Factor the standard form $3x^2-30x+48$.
First factor out 3: $y=3(x^2-10x+16)$
Factor quadratic: $x^2-10x+16=(x-2)(x-8)$
So $y=3(x-2)(x-8)$

Answer:

	Standard Form	Vertex Form	Factored Form
2	$y=2x^2-8x-24$	$y=2(x-2)^2-32$	$y=2(x-6)(x+2)$
3	$y=4x^2-24x-28$	$y=4(x-3)^2-64$	$y=4(x-7)(x+1)$
4	$y=5x^2-30x+40$	$y=5(x-3)^2-5$	$y=5(x-2)(x-4)$
5	$y=5x^2-20x-225$	$y=5(x-2)^2-245$	$y=5(x-9)(x+5)$
6	$y=3x^2-30x+48$	$y=3(x-5)^2-27$	$y=3(x-2)(x-8)$