Question

Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.

Answer 1

Ok, tell me this:

Say, you have: \(x^2-8x\)
THEN, what number would you want to add to the end,
to make that a perfect square trinomial?

Answer 2

8? I think

Answer 3

probably not

Answer 4

\(\large\color{black}{ \displaystyle x^2\pm\color{red}{\rm 2a}x+\color{red}{\rm a}^2 }\)

Answer 5

Does this look familiar to anything?

Answer 6

`(x - a)² = x² - 2a + a²` `(x + a)² = x² + 2a + a²`

Answer 7

umm no. I learned this but can't get the answer

Answer 8

wait i know that

Answer 9

you know those two in gray, right?

Answer 10

sum and difference of a cube??

Answer 11

No cubes:

Just how to expand (x+a)^2 and (x-a)^2

Answer 12

ohok

Answer 13

So, we are going to make x^2-8x (by adding some number)

into x^2-2a+a^2 form

Answer 14

ok

Answer 15

\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 2a}x+\color{red}{\rm a}^2 =(x-\color{red}{\rm a})^2 }\)

Answer 16

So if 2a is 8, than a is 4. Right?

Answer 17

ya

Answer 18

than you square it?

Answer 19

\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x+\color{red}{\rm 4}^2 =(x-\color{red}{\rm 8})^2 }\)

\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x+16 =(x-\color{red}{\rm 8})^2 }\)

that would be a perfect square trinomial.
So we add 16 to the x^2-8x, to make our polynomial a perfect square.

Answer 20

what is this formula called? cause I didn't learn this?

Answer 21

which formula?

Answer 22

the one you are using

Answer 23

oh, don't worry, you can just google (a+b)^2 and all of that will come up

Answer 24

all of those like (a+b)^2 and (a-b)^2 ... it will all show up.

Answer 25

okk

Answer 26

We will add 16, but we can NOT just add 16, because we will change
the value of the trinomial.

So this is what we will do:

\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{}+3 }\)

\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{+16-16}+3 }\)

\(\large\color{black}{ \displaystyle (x^2-\color{red}{\rm 8}x+16)-16+3 }\)

in parenthesis, it is a perfect square trinomial
and -16+3 is -13.

Answer 27

so it would be -8-13+3

Answer 28

you mixed up something a little, how did you get that result/

Answer 29

??

Answer 30

from that equation

Answer 31

wait x- 8 - 13?

Answer 32

\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 2a}x+\color{red}{\rm a}^2 =(x-\color{red}{\rm a})^2 }\)

And we want to get our \(x62-8x\) into that form, by adding a value.
We will add 16, but we can NOT just add 16, because we will change
the value of the trinomial.

So this is what we will do:

\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{}+3 }\)

\(\large\color{black}{ \displaystyle x^2-\color{red}{\rm 8}x\color{blue}{+16-16}+3 }\)

\(\large\color{black}{ \displaystyle (x^2-\color{red}{\rm 8}x+16)-16+3 }\)

I added a magic zero, so to speak.
+16-16, is all I did:)

in parenthesis, it is a perfect square trinomial
and -16+3 is -13.

\(\large\color{black}{ \displaystyle (x^2{~}-{~} 2\times\color{red}{ 4}x{~}+{~}\color{red}{ 4}^2)-13 }\)

do you see that x\(^2\)-2a+a\(^2\) form in the parenthesis?
We know that `a²-2a+a²=(x-a)²`, and thus we get the following:

\(\large\color{black}{ \displaystyle (x-\color{red}{ 4})^2-13 }\)

Answer 33

read it carefully, and if you have questions, then ask.

Answer 34

ohok tysm!! so you just simplied 8 to 4?

Answer 35

oh, the result I showed at the end can't be simplified.
You can re-write it to what it was in the beginning if you do the inverse operations.

Answer 36

ohhhok f(x) = (x − 4)2 − 13 this would be the answer right?

Answer 37

(x-4)\(\color{red}{^2}\)-13 :D

Answer 38

tysm! can you hep with other one on new thread plz?