Question

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.

standard form

vertex form

factored form

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1

$y=(x - 2)(x + 6)$

2

$y=-2x^2 - 8x - 6$

3

$y=(x + 1)^2 - 16$

4

$y=2x^2 - 20x + 48$

5

$y=2(x - 4)(x - 2)$

6

$y=2(x - 2)^2 - 32$

Explanation:

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Problem 1: Given Factored Form $y=(x-2)(x+6)$

Step1: Expand to standard form

Multiply the binomials:

$$\begin{align*} y&=(x-2)(x+6)\\ &=x^2+6x-2x-12\\ &=x^2+4x-12 \end{align*}$$

Step2: Complete the square for vertex form

Isolate $x$-terms, add/subtract $(\frac{4}{2})^2=4$:

$$\begin{align*} y&=x^2+4x-12\\ &=(x^2+4x+4)-4-12\\ &=(x+2)^2-16 \end{align*}$$

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Problem 2: Given Standard Form $y=-2x^2-8x-6$

Step1: Factor out leading coefficient

Factor $-2$ from $x$-terms:
$$y=-2(x^2+4x)-6$$

Step2: Complete the square for vertex form

Add/subtract $(\frac{4}{2})^2=4$ inside the parentheses:

$$\begin{align*} y&=-2(x^2+4x+4-4)-6\\ &=-2((x+2)^2-4)-6\\ &=-2(x+2)^2+8-6\\ &=-2(x+2)^2+2 \end{align*}$$

Step3: Convert to factored form

Set $y=0$, solve for $x$:

$$\begin{align*} -2(x+2)^2+2&=0\\ (x+2)^2&=1\\ x+2&=\pm1\\ x&=-1 \text{ or } x=-3 \end{align*}$$

Write as factored form:
$$y=-2(x+1)(x+3)$$

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Problem 3: Given Vertex Form $y=(x+1)^2-16$

Step1: Expand to standard form

Expand the squared binomial:

$$\begin{align*} y&=(x^2+2x+1)-16\\ &=x^2+2x-15 \end{align*}$$

Step2: Factor to factored form

Find two numbers that multiply to $-15$ and add to $2$ ($5$ and $-3$):
$$y=(x+5)(x-3)$$

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Problem 4: Given Standard Form $y=2x^2-20x+48$

Step1: Factor out leading coefficient

Factor $2$ from all terms:
$$y=2(x^2-10x+24)$$

Step2: Complete the square for vertex form

Add/subtract $(\frac{-10}{2})^2=25$ inside the parentheses:

$$\begin{align*} y&=2(x^2-10x+25-25+24)\\ &=2((x-5)^2-1)\\ &=2(x-5)^2-2 \end{align*}$$

Step3: Convert to factored form

Factor the quadratic inside the parentheses (numbers $-6$ and $-4$ multiply to $24$, add to $-10$):
$$y=2(x-6)(x-4)$$

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Problem 5: Given Factored Form $y=2(x-4)(x-2)$

Step1: Expand to standard form

Multiply binomials first, then distribute $2$:

$$\begin{align*} y&=2(x^2-2x-4x+8)\\ &=2(x^2-6x+8)\\ &=2x^2-12x+16 \end{align*}$$

Step2: Complete the square for vertex form

Factor $2$ from $x$-terms, add/subtract $(\frac{-6}{2})^2=9$:

$$\begin{align*} y&=2(x^2-6x)-16\\ &=2(x^2-6x+9-9)+16\\ &=2((x-3)^2-9)+16\\ &=2(x-3)^2-18+16\\ &=2(x-3)^2-2 \end{align*}$$

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Problem 6: Given Vertex Form $y=2(x-2)^2-32$

Step1: Expand to standard form

Expand the squared binomial, distribute $2$:

$$\begin{align*} y&=2(x^2-4x+4)-32\\ &=2x^2-8x+8-32\\ &=2x^2-8x-24 \end{align*}$$

Step2: Convert to factored form

Factor out $2$, then factor the quadratic (numbers $-6$ and $2$ multiply to $-12$, add to $-4$):

$$\begin{align*} y&=2(x^2-4x-12)\\ &=2(x-6)(x+2) \end{align*}$$

Answer:

	Standard Form	Vertex Form	Factored Form
2	$y=-2x^2-8x-6$	$y=-2(x+2)^2+2$	$y=-2(x+1)(x+3)$
3	$y=x^2+2x-15$	$y=(x+1)^2-16$	$y=(x+5)(x-3)$
4	$y=2x^2-20x+48$	$y=2(x-5)^2-2$	$y=2(x-6)(x-4)$
5	$y=2x^2-12x+16$	$y=2(x-3)^2-2$	$y=2(x-4)(x-2)$
6	$y=2x^2-8x-24$	$y=2(x-2)^2-32$	$y=2(x-6)(x+2)$