Question

1. What is the vertex form of the equation?

Answer 1

1.vertex form is $y=a(x-h)^2+k$

to get into vertex form, complete the square

$y=-x^2+12x-4$

step 1: group x terms

$y=(-x^2+12x)-4$

step 2: factor coefient in front of x^2 term

$y=-1(x^2-12x)-4$

prep for next step: take 1/2 of linear (1st dgree term) and square it

$\frac{-12}{2}=-6$, $(-6)^2=36$

step 3: add positive and negative of that previous number inside the parenthasees

$y=-1(x^2-12x+36-36)-4$

factor perfect squaer trionmial

$y=-1((x-6)^2-36)-4$

take out that -36 by expansion

$y=-1(x-6)^2+36-4$

$y=-1(x-6)^2+32$

2. factor out the 2

$2(x^2+8x+12)$

what 2 numbers multiply to get 12 and add to get 8?

2 and 6

$2(x+2)(x+6)$

3.

for $ax^2+bx+c=0$

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

in our case

$x^2-7x-6=0$

a=1, b=-7, c=-6

$x=\frac{-(-7) \pm \sqrt{(-7)^2-4(1)(-6)}}{2(1)}$

$x=\frac{7 \pm \sqrt{73}}{2}$

so $x=\frac{7+\sqrt{73}}{2}$ and $x=\frac{7-\sqrt{73}}{2}$

4. remember that $i=\sqrt{-1}$ so $i^2=-1$

treat $i$ as a variable

$(-2i)(8i)=$

$(-2)(i)(8)(i)=$

$(-2)(8)(i)(i)=$

$(-16)(i^2)=$

$(-16)(-1)=$

[tex}16[/tex]

Answer 2

1. Vertex form

y = - (x^2 - 12 x +     ) -4 +    

Complete the square

y = - (x^2 - 12 x +  36 ) -4 + 36

y = - (x - 6)^2 + 32

2. Factored form

Take out 2 from the equation like so:

2(x^2 + 8x + 12)

2 + 6 = 8

2(x+2)(x+6)

3. Quadratic Formula

x = −b± √b^2 −4ac over

2a

Plug

7​ ± √49 + 24 over

2

7   ± √73 over

2

4. Simplify

-16 i^2

Since i^2 is equal to -1 do:

(-16) (-1) = 16

16

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