Question

Can anybody help me find the vertex for y=2x^2-4x? Please explain answer.

Answer 1

Complete the square to put it in vertex form. Easy peasey. Or you can find your 0's (since this one has them) and the vertex will be half-way between them.

Answer 2

???

Answer 3

I have a formula for it.

Answer 4

Clouser, are you familiar with completing the square?

Answer 5

Which part do you not understand?

Answer 6

completing the square or finding the 0's?

Answer 7

So You make it y=2x(x-4)

Answer 8

Or sorry y=2x(x-2)

Answer 9

No take 2x^2-4x
Factor out the 2.
\[y=2(x^2-2x)\]
\[y=2(x^2-2x+1-1)\]
\[y=2(x-1)^2-2\]
So the vertex is (1,-2)

Answer 10

vertex form is
\[y=a(x-h)^2+k\]
(h,k) is vertex

Answer 11

He was finding the 0s which is ok too

Answer 12

No use myiniaya's information to read it off my algebra.

Answer 13

Since this one has them

Answer 14

lol

Answer 15

What do you mean by finding the 0's?

Answer 16

The two 0's you have there Clouse are 0 and 2. So the vertex will be at x is halfway between those.

Answer 17

what is wrong with normal old white bread and butter
\[-\frac{b}{2a}\]?

Answer 18

y = 2x(x-2) means that y = 0 when x is 0 or when x is 2

Answer 19

\[\frac{0+2}{2}\]

Answer 20

-b/2a is too hard to try to remember ;p

Answer 21

yeah satellite

Answer 22

its too hard to remember

Answer 23

good lord. just use
\[-\frac{b}{2a}=\frac{4}{4}=1\]

Answer 24

lol

Answer 25

No! You can't make me! ;p

Answer 26

Formula for h is -b/2a and formula for k is c-h^2a. :)

Answer 27

My brain has only enough room for 1 forumla involving quadratic polynomials. So I am using it for the \[x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] one.

Answer 28

yes i know it is very hard to remember. much harder than completing the square . it has the following hard symbols
\[-\]
\[b\]
/

\[2\]
\[a\]

Answer 29

Everything else I derive.

Answer 30

LOL, thanks guys. Just discovered this site today. It is like having lots of tutors.

Answer 31

Sorry for the tangent Clouser.

Answer 32

very difficult. must easier to factor and keep track of what you added and subtracted to complete the square. or take the derivative, set equal zero and solve, or find the point between the two roots. but it is still
\[-\frac{b}{2a}\]

Answer 33

Satellite, from this equation how do you find the axis of symmetry?

Answer 34

and no matter what you do you are going to end up with
\[(-\frac{b}{2a},f(-\frac{b}{2a}))\]

Answer 35

\[y=ax^2+bx+c=a(x^2+\frac{b}{a}x)+c=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2\]
\[y=a(x+\frac{b}{2a})^2+c-\frac{b}{4a}\]
\[y=a(x-(-\frac{b}{2a}))^2+(c-\frac{b}{4a})\]
vertex is
\[(-\frac{b}{2a},c-\frac{b}{4a})\]

Answer 36

it is
\[y=-\frac{b}{2a}\] how hard is that?

Answer 37

So in a nutshell there are 3 ways to find the vertex.

1) \(\large x_{vertex} = -\frac{b}{2a}\)
2) Re-write in vertex form.
3) Find the midpoint between the 0's of the quadratic (if it has them)

Answer 38

and pretty clearly method one is easiest

Answer 39

Thanks so much.

Answer 40

we all say you are welcome

Answer 41

and we all like to fuss with each other. it is fun

Answer 42

my is clearly the easiest
to derive it each time

Answer 43

lol

Answer 44

oh yes. actually easiest is finding the derivative using the difference quotient. on an etchasketch

Answer 45

easiest to do sure. easiest to remember, nope.

The last one is easiest to remember (most intuitive), but not always reliable.
The first one is easiest to do if you remember it.
The middle one is a nice middle ground and always works.

Answer 46

minus bee over two ay?

Answer 47

man i thought my memory was bad...

Answer 48

totally arbitrary data.

Answer 49

like polpak said its hard too remember lol

Answer 50

oh but it is not arbitrary at all. it is what you get

Answer 51

satellite you just need to conform to my way

Answer 52

putting it in vertex form is the bestest
i win!

Answer 53

\[y=ax^2+bx+c\]
\[y=a(x+\frac{b}{2a})^2+\text{stuff}\]
vertex is clearly when
\[x=-\frac{b}{2a}\] because that makes the perfect square part zero

Answer 54

I agree.

Answer 55

+stuff?

Answer 56

Vertex form is best.

Answer 57

yeah stuff. i don't feel like keeping track. if i want "stuff" i replace x by
\[-\frac{b}{2a}\] so see what i get

Answer 58

what could be easier?

Answer 59

See I already screwed it up in this coversation. I thought it was -a/2b

Answer 60

who want to say "i added this but i really added that so i have to subtract the other"?

Answer 61

Totally arbitrary.

Answer 62

i want to say that

Answer 63

You probably like foil too ;p

Answer 64

@poplak look at this and tell me why it is arbitrary instead of obviously true
\[y=ax^2+bx+c\]
\[y=a(x+\frac{b}{2a})^2+\text{stuff}\]

Answer 65

first, outer, inner, last, what could be easier?

Answer 66

gimme a break there is a reason for it to be
\[-\frac{b}{2a}\] foil is for turkeys at thanksgiving

Answer 67

Because you're completing the square

Answer 68

Which is basically solving for vertex form.

Answer 69

Except you jump to the end and remember some formula -a/2b or some such.

Answer 70

\[y=ax^2+bx+c=a(x^2+\frac{b}{a}x)+c=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2\]
\[y=a(x+\frac{b}{2a})^2+c-\frac{b}{4a} \]
\[y=a(x-(-\frac{b}{2a}))^2+(c-\frac{b}{4a}) \]
satellite why don't you just remember that the vertex is just
(-\frac{b}{2a},c-\frac{b}{4a})
why do you only like to remember the x part only?

Answer 71

You guys are still here...LOL

Answer 72

\[(-\frac{b}{2a},c-\frac{b}{4a}) \]

Answer 73

because i find it easy to evaluate a quadratic funciton

Answer 74

NERD FIGHT!

Answer 75

Commence flailing and hand slapping on the count of three!

Answer 76

but at least you must agree that when you complete the square, not matter how many times you do it, the first coordinate will be
\[-\frac{b}{2a}\] so there is nothing at all arbitrary about it

Answer 77

but
\[f(-\frac{b}{2a})=c-\frac{b}{4a}\]
whats the point in pluggin' that if you are just gonna get what i got

Answer 78

But you have to remember it that way instead of remembering 'complete the sqaure to get vertex form'. I can remember methods. I cannot remember formula.

Answer 79

why would you want to repeat the procedure again and again. unless you are trying to teach someone how to do it

Answer 80

why don't you remember everything satellite since remembering is so much easier

Answer 81

Dyslexia is a feather I guess is what it boils down to.

Answer 82

@polpak i do not actually believe you because if you wanted to solve
\[3x^2+5x-1=0\] i am willing to bet that you will NOT complete the square

Answer 83

frock together

Answer 84

\[\text{feather} = \text{b}\text{i}\text{t}\text{c}\text{h}\]

Answer 85

i would use my new method to factor that
and i'm havent shown polpak my new method

Answer 86

really? what language?

Answer 87

In the silly anti-cursing filter they have here.

Answer 88

assumption

Answer 89

they seem to have fixed it

Answer 90

pass the class

Answer 91

don't be an retrice

Answer 92

oh they fixed it

Answer 93

apparently.

Answer 94

what did they fix

Answer 95

good because it was asinine

Answer 96

In any event, I'm dyslexic. So I cannot remember random jumbles of letters and numbers in formula. I have to remember methods.

Answer 97

oh you couldn't type any word with
retricein it

Answer 98

oh they really fixed it