Question

what is the vertex form of the equation
y=-x^2 +12x -4

Answer 1

\(\bf \textit{vertex of a parabola}\\ \quad \\
y = {\color{red}{ 1}}x^2{\color{blue}{ +12}}x{\color{green}{ -4}}\qquad

\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\)

Answer 2

hmm actually, you're only asked for the form anyhow

have you covered what a "perfect square trinomial" is?

Answer 3

or also called 'completing the square'

Answer 4

no, not yet

Answer 5

hmmm ok minimoo -> \(\large \begin{array}{cccccllllll}
{\color{brown}{ a}}^2&\pm &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}}\pm {\color{blue}{ b}})^2&\leftarrow
&&

\end{array}\)

looks familiar?

Answer 6

not really

Answer 7

ok..... well

notice the "a" and "b"

if you had a trinomial with two terms are squared
and another one.. .is "twice" the non-squared form of the other two
is a perfect square trinomial

notice the form, notice the middle term, is just 2 * the other two, without the exponent

Answer 8

okay

Answer 9

so... now let's take a look at your trinomial
so \(\bf -x^2 +12x -4\implies (-x^2 +12x )-4\implies -(x^2-12x)-4
\\ \quad \\
-(x^2-12x+{\color{red}{ \square ?}}^2)-4\)

what number do you think we could use there, to have a perfect square trinomial in the parentheses?

Answer 10

3?

Answer 11

3....ok... let us test that, recall that \(\bf {\color{brown}{ a}}^2\pm 2{\color{brown}{ a}}{\color{blue}{ b}}+{\color{blue}{ b}}^2\)

so that means 12x or your middle term
would be 2 * x * "unknown"

so you said unknown = 3

let's see

12x = 2 * x * 3
12x = 2x * 3
12x\(\ne\) 6x <-- well.. is not 3 then

Answer 12

oh.... it would be 6 to make it equal. i was thinking 12 x 3 which would be 36 which is a perfect square

Answer 13

yeap
12x = 2 * x * 6
12x = 12x <--- so yes, it's 6

recall \(\large \begin{array}{cccccllllll}
{\color{brown}{ a}}^2&\pm &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}}\pm {\color{blue}{ b}})^2&\leftarrow
&&

\end{array}\)

keep in mind that all we're doing is "borrowing from our good friend 0"
so if we ADD 6, we also have to SUBTRACT 6

thus

\(\bf -x^2 +12x -4\implies (-x^2 +12x )-4\implies -(x^2-12x)-4
\\ \quad \\
-(x^2-12x+{\color{red}{ 6}}^2-{\color{red}{ 6}}^2)-4\implies -(x^2-12x+{\color{red}{ 6}}^2)+{\color{red}{ 6}}^2-4\)

so.. .can you take it from there?

Answer 14

well... we're really adding and subtracting \(\bf 6^2\) to be clear

Answer 15

but then how would i get it to change to the equation y= a(x-h)^2 +k

Answer 16

\(\bf -x^2 +12x -4\implies (-x^2 +12x )-4\implies -(x^2-12x)-4
\\ \quad \\
-(x^2-12x+{\color{red}{ 6}}^2-{\color{red}{ 6}}^2)-4\implies -(x^2-12x+{\color{red}{ 6}}^2)+{\color{red}{ 6}}^2-4
\\ \quad \\
-[{\color{brown}{ x}}^2-2({\color{brown}{ x}})({\color{blue}{ 6}})+{\color{blue}{ 6}}^2]+6^2-4\)

what do you think? can we simplify the parentheses group?

Answer 17

well... bracketed rather =)

Answer 18

woah. i got lost when both 6^2 turned positive

Answer 19

notice the " - "
in front of the parentheses -> \(\bf -(x^2-12x+{\color{red}{ 6}}^2-{\color{red}{ 6}}^2)-4\implies -(x^2-12x+{\color{red}{ 6}}^2)+{\color{red}{ 6}}^2-4\)

if you take any of the group terms out... you'd have multiply it by that

Answer 20

oh okay and then after that what would the splitting of the 12x to 2(x)(6) do/

Answer 21

\(\large \begin{array}{cccccllllll}
{\color{brown}{ a}}^2&\pm &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}}\pm {\color{blue}{ b}})^2&\leftarrow
&&

\end{array}\)

so.... any ideas on simplifying the bracketed group?

Answer 22

(x +- 6)^2

Answer 23

yes.... in this case we have a minus for the middle term, or 12x
so it'd be \((x-6)^2\ then
thus \(\bf -x^2 +12x -4\implies (-x^2 +12x )-4\implies -(x^2-12x)-4
\\ \quad \\
-(x^2-12x+{\color{red}{ 6}}^2-{\color{red}{ 6}}^2)-4\implies -(x^2-12x+{\color{red}{ 6}}^2)+{\color{red}{ 6}}^2-4
\\ \quad \\
-[{\color{brown}{ x}}^2-2({\color{brown}{ x}})({\color{blue}{ 6}})+{\color{blue}{ 6}}^2]+6^2-4\implies -(x-6)^2+36-4
\\ \quad \\
-(x-6)^2+32\)

Answer 24

how can you tell when you have simplified enough?

Answer 25

when you can't "simplify" it more

36-4 simplified to 32

and the trinomial, simplifed to a binomial

and that'd be the vertex form of the parabola

Answer 26

\(\large {
\begin{array}{llll}

y=(x-{\color{brown}{ h}})^2+{\color{blue}{ k}}\\
x=(y-{\color{blue}{ k}})^2+{\color{brown}{ h}}
\end{array}
\qquad
\begin{array}{llll}
vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\

\end{array}}\)

which means... based on that vertex form, the vertex of that parabola is at ( 6, 32)

thus is called the "vertex form" of the equation

Answer 27

ahhh x) okay.

Answer 28

thanks soo much you were such a great help :)

Answer 29

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIteF4yKzEyeC00IiwiY29sb3IiOiIjQkYyNjI2In0seyJ0eXBlIjoxMDAwLCJ3aW5kb3ciOlsiLTIuMDQ0Mjk4NTI5NjI0OTQwNyIsIjE4LjI2ODIwMTQ3MDM3NTA0IiwiMjMuMjM4MDM0NTU4Mjk2MTgiLCIzNS43MzgwMzQ1NTgyOTYxOCJdfV0-

leading term is negative... so it opens downward
yw

Answer 30

:)