# Homework Help: Find vertex by completing the square-NASTY

1. Jul 22, 2007

### sakin

1. The problem statement, all variables and given/known data
find the vertex by completing the square:

f(x)=2x^2-4x+7

2. Relevant equations

3. The attempt at a solution

numerous attempts yet not one looks right.

one was as such:

y-7=2x^2-4x;
y-7+4=2x^2-4x+4;
y-3=2x^2-4x+4;

y=2(x-2)^2 +3... I could see this as (2,3) if I didn't have that nasty 2 in front of the square.

so

I tried

y/2-7/2=x^2-2x;
y/2-7/2+2=x^2-2x+2;
y/2-3/2=(x-2)^2;

y/2=(x-2)^2+3/2... so it could be (2, 3/2) but then there's a denominator to deal with, and then I'd be right back where I started.

anyone here make some sense of this? THX

2. Jul 22, 2007

### rocomath

so you have 2x^2 - 4x + 7

you can start by factoring out a 2 or dividing by 2. *factor by grouping*

on a side note, the vertex of a parabola is simply the average of its two solutions.

Last edited: Jul 22, 2007
3. Jul 22, 2007

### sakin

well the question asked for vertex form, and it turns out I had vertex form... it's just with an 11-week course it's all too easy to forget a bunch of stuff.

y=2(x-2)^2+3... looks like a vertex to me.

how many points do I get for fixing my own problem? :rofl:

Last edited: Jul 22, 2007
4. Jul 22, 2007

### rocomath

i have a detailed outline of how i did the problem, could you please show me your steps to how you simplified your equation b4 i show you.

i took a 6-week Calculus course and i remember everything. you have to work at it! one of the best things in math is the fact that everything you learn, there is probably a way to apply it to what you're currently learning. anyways, your answer is way off or i suck.

Last edited: Jul 22, 2007
5. Jul 22, 2007

### sakin

y-7=2x^2-4x;
y-7+4=2x^2-4x+4;
y-3=2x^2-4x+4;

y=2(x-2)^2 +3

just as I had done it in my first attempt, original posting.

6. Jul 22, 2007

### rocomath

i went over my work twice, hopefully i made no errors. so what is the answer?

Last edited: Jul 22, 2007
7. Jul 22, 2007

### sakin

with all due respect, partner, you're spinning your wheels in gravel.

the problem was to put the quadratic into vertex form by completing the square, vertex form being

y= a(x-h)^2+k

and I did that via

f(x)=2(x-2)^2+3

8. Jul 22, 2007

### rocomath

http://img410.imageshack.us/img410/7213/vertexzm1.jpg [Broken]

Last edited by a moderator: May 3, 2017
9. Jul 22, 2007

### Gokul43201

Staff Emeritus
Check your math: 2(x-2)^2 != 2x^2-4x+4

10. Jul 22, 2007

### cristo

Staff Emeritus
If you expand your answer, sakin, you don't get the original expression.

Here's how I'd do it:
Firstly, factor out the 2 to obtain 2(x^2-2x)+7.
Now, complete the square in the brackets 2[(x-1)^2-1]+7
Expand and collect 2(x-1)^2+5.

11. Jul 22, 2007

### sakin

folks, this is a COLLEGE ALGEBRA course... not calculus.

The question asked to put the equation in vertex form by completing the square.

f(x)=2X^2-4X+7

now I set f(x) to y. I know you purists out there probably hate to see this but bear with me, I'm not, repeat NOT a math aficionado.

I move the 7 over to the other side:
y-7=2X^2-4X

or if you like:

2X^2-4X=y-7

then:

2X^2-4X +(-4/2)^2=y-7+(-4/2)^2.

so now it's 2X^2-4X+4=y-3

so to conclude things:

f(x)=2(X-2)^2+3.

is this, or is this not, the vertex form of the original equation?

for the record, I graphed the original equation, and then all suggested answers- each was a different graph. I really believe this to be a conversion problem and nothing more. FWIW, the teacher makes this stuff up on the fly.

I am not concerned with finding zeroes or any such extra steps. I appreciate some of the rather excruciating work undertaken in above postings, but it was not necessary given what the problem asked.

one thing I know about math now, is that it is a proof of a quote I believe was attributed to Mark Twain:

"there are lies, there are damned lies, and then there are statistics."

Last edited: Jul 22, 2007
12. Jul 22, 2007

### cristo

Staff Emeritus
Noone's saying it is, and if you read my post, I don't use any calculus, simply correct one of your mistakes! You need to factor out the two before you can continue with completing the square.

13. Jul 22, 2007

### cristo

Staff Emeritus
You obviously did it incorrectly, as 2(x-1)^2+5=2x^2-4x+7

14. Jul 22, 2007

### sakin

mistyped a character- my mistake.

so can someone explain, in layman's terms, how one concludes that 2(x-1)^2+5 is the solution?

here is my beef- your process does not jibe with the the way the examples are worked in the book. I don't see the 7 being moved over to the other side in the Rockswold text. So having everything on the other side is not what a non-math person wants to see when the examples in the text, Purplemath.com, West Texas A&M, etc. move stuff over when completing the square so that all that is left is ax^2+bx to work with.

I will make one subjective statement in all of this- I find that, like the brutal math teacher, you folks are more geared to the math-minded. A more patient tone with people desperately seeking help would be appreciated.

Last edited: Jul 22, 2007
15. Jul 22, 2007

### cristo

Staff Emeritus
What do you not understand about the working in post #10? If you're not gonna read what I'm writing then there's not really much point in me writing it is there?

16. Jul 22, 2007

### rocomath

did you even look at how i worked the problem? my god ... i wrote out every single step, it is exactly what you are looking for. you need to accept that you worked it out incorrectly, and the fact that there are different ways to arrive at the answer. we're being as helpful and patient as we can be.

17. Jul 22, 2007

### sakin

see my above posting. This may make perfect sense to you but I'm still trying to follow your formula and getting nowhere.

2[(x-1)^2-1]+7

becomes

2[x^2-2x+1-1]+7

which I see as

2(x^2-2x)+7

so what happened? where did a -2 outside of the parenthesis come from? Can you explain to a non-mathematician?

Last edited: Jul 22, 2007
18. Jul 22, 2007

### rocomath

in order to complete my square, i have to take the -1 out by multiplying it by the coefficient 2, which became -2.

a simple way of getting the Vertex is through the equation -b/2a which gives me my x-value, then plug in that x-value into your equation which gives you your y-value.

i'm not sure what your financial stand-point is but, this is by far the best textbook i have ever used. the explanations are extremely easy to understand and its geared towards a prep for Calculus. though idk if you're going into Calculus, it's still a great text that really explains Algebra well. i purchased it for used for \$20. i was struggling a lot with Calculus the first week bc my Algebra sucked so bad, but damn this book saved my life :-]

Algebra and Trigonometry (2nd Edition) (Beecher/Penna/Bittinger Series) - https://www.amazon.com/Algebra-Trig...3546315?ie=UTF8&s=books&qid=1185158249&sr=8-4

and if you're going into Trigonometry, you have 2 in 1!

Last edited by a moderator: Apr 22, 2017
19. Jul 22, 2007

### cristo

Staff Emeritus
Well it is 2(x^2-2x)+7; but you've worked backwards away from the right answer!!

OK, let's try again, using slightly different notation we have
y=2x^2-4x+7
Now, rearranging gives y-7=2x^2-4x
divide both sides by 2: 1/2(y-7)=x^2-2x

Now comes the trick on the RHS. You can either add and subtract 1 from the RHS (which, to me, seems counterintuative). Or, you can follow the procedure. Halve the coefficient of x and write it in the brackets (x-1)^2, then subtract the square of the constant coefficient. (-1)^2=1, hence we obtain
1/2(y-7)=(x-1)^2-1

Rearranging gives y=2(x-1)^2-2+7=2(x-1)^2+5.

Does this help? If not, then you should probably ask your teacher, as it's a lot easier to explain this in person than over the internet!!

20. Jul 22, 2007

### sakin

see where one could find this both arbitrary AND contradictory? I combined like terms at [x^2-2x+1-1], where 1-1 removes that term, and now I find that I wasn't supposed to follow that rule in this case?

Again, you may find this as elemental as tying your shoes but I obviously don't. I find that tutorials are largely insufficient so I come in here and the implication is that I'm a dimwit.