Find vertex by completing the square-NASTY

[sakin]

Homework Statement

find the vertex by completing the square:

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

Homework Equations

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

 

[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.

 

[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:

 

[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.

 

[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.

 

[rocomath]

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

 

[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

 

[rocomath]

http://img410.imageshack.us/img410/7213/vertexzm1.jpg

 

[Gokul43201]

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

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

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

 

[cristo]

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.

 

[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."

 

[cristo]

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

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.

 

[cristo]

for the record, I graphed the original equation, and then all suggested answers- each was a different graph.

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

 

[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.

 

[cristo]

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

 

[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.

 

[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?

 

[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/dp/0321159357/?tag=pfamazon01-20

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

 

[cristo]

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? Can you explain to a non-mathematician?

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!

 

[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.

 

[rocomath]

one of the steps in completing the square is by adding the number that you get by dividing bx by 2 and then squaring it to both sides. but the thing is, we squared within a parenthesis so we kept the number within the parenthesis. by doing 1-1, we're implying that we're adding 0, which is basically completing the square w/o changing the equation.

i don't think you're a dimwit, hell, i didn't know any of this **** beginning of summer. i worked really hard this summer to bring myself back to speed bc i lacked so much mathematical foundation. i hope you're not discouraged, we're human too. we're trying to explain it the best we can, it's frustrating on both ends. all love tho :-]

 

[sakin]

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

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!

Problem is, the teacher has even LESS patience than what I have seen in darker moments here. He's ram-rodding through an 11-week summer course and really has no patience for any of us trying to get him to explain problems based on concepts he feels he already covered. So again, that's why I'm here. He acts like he should be at MIT, and I'm just trying to get through this class as a pre-req for health professions.

 

[cristo]

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?

I see what you mean, but the reason that the "1-1" is included in the above is precisely so that it can be factored to (x-1)^2-1.

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.

sakin, no-one's calling you a dimwit, and I certainly am not meaning to imply anything of the sort. It's a fact of life that there are some things that are incredibly difficult for one to understand, no matter how many times they are told. In my education thus far, there have been various occasions where I simply did not understand something, no matter how many times I tried it, or tried to read how to do it. However, the majority of these problems have been solved by asking a teacher to go through something one on one with me. There is certainly no reason to feel stupid for doing this! Physicsforums is a very useful place for homework and general questions about learning, but I think that there are some things that one simply has to be sat in the same room as a teacher, reading what the teacher writes as he puts it on the board, in order to understand. I've also tutored people through high school classes, and the mere fact of being there pointing to the paper whilst describing a technique to someone enables them to understand it. That's why I think it best you ask your teacher to do this-- not because I think you're stupid!

Of course, I'll be ready to bet that someone will come along and explain this is a far more succinct manner than I have, and enable you to understand it, but then that'll be good for you won't it!

 

[sakin]

I appreciate the help, really I do. I'm just trying to get into a sonography program, take two classes this summer, get ready for fall term and still raise two daughters and try to be a good wife. I realize I'm paying now for majoring in registration with a minor in drops back in the day but I'm serious now.

 

[symbolipoint]

Sakin,
The goal when using "completion of the square" is to transform an expression into "standard form", which renders finding a vertex point to be easy to do. When you complete the square, you also may need to undo the process - why? ... to maintain an equivalent expression (which appears different but means the same as before).

NEVER BE AFRAID to study the same topic more than once. Most students & former students who studied Intermediate Algebra struggled for at least a few weeks to learn about completing the square, but with extra practice and repeated study, they become comfortable and skillful with it. Your 11 week summer course may be stuffed into that time span from an ordinary term length of 16 or 18 weeks. The condensed time given to the summer session makes the course potentially more difficult if this is your first time studying the material.

You could try an internet search to find a graphical representation of competing the square in case your textbook does not show one.

 

[Tedjn]

I'll give my shot at explaining this as well (it will help me understand this better also). The following is a long ramble (I tend to do that). If it's something you already understand, then just ignore it, since I don't know how much you know.

As symbolipoint said, when we "complete the square," we are really changing the quadratic equation from some form/representation to an equivalent representation mathematicians call standard form. It is important to realize that the goal is to transform the quadratic equation into standard form, and it doesn't really matter how you do it. "Completing the square" is perhaps the easiest method, but never forget that it is really just a tool to get your equation into standard form.

First of all, why would we want to transform the quadratic equation into standard form? One of the forms of the quadratic equation we are often taught is $$f(x)=y=ax^2+bx+c$$, but it is not clear from the equation that the graph should be a parabola. We are told by our teachers that the graph is a parabola, but why?

Take an example. Suppose our equation is $$y=x^2+4x+1$$. We graph this equation with several points, and the more points we add, the more we see that, indeed, the graph does look like it is a parabola. But why? We don't see this intuitively. Why does one x^2 plus a 4x plus a 1 give me a parabola? What if I had $$x^2+4x+2$$? Would that still be a parabola? What if I had $$5x^2+33x+43$$? Would that still be a parabola?

We try several more sample equations, and we find that all equations in the form $$y=ax^2+bx+c$$ seem to be parabolas. We have a strong suspicion that they are all parabolas, but we still can't see why from the equation. Let's play around with the equation to see if we can make it look like something we would associate with a parabola.

Take $$y=x^2+4x+1$$ again. We can't factor this equation. After staring at it form awhile, however, we may get a bright idea. What if we add something to both sides to make the right hand side something that can be factored? Let's make it easy by just making the right hand side a perfect square. We recall that $$(x+a)^2 = x^2+2ax+a^2$$. From our equation, it looks like $$2a=4 \implies a=2$$. Then the constant term $$a^2=4$$. Right now, we have a 1, but we want to make it a 4. We can add 3 to both sides, however. Then, we would have

$$y+3 = x^2 + 4x + 4$$

You should know that we chose the right hand side to be a perfect square completely arbitrarily. But you'll also see that we were lucky that we made this choice, because it will soon enable us to see why we have a parabola.

This is also why the method is called "completing the square," because we are making a perfect square on the right hand side. Now, you might be thinking, my book would have told me to first subtract 1 on both sides. I would then have

$$y-1 = x^2+4x \implies y-1+4 = x^2+4x+4$$,

but can you see that this is really the same thing? We are really adding three to both sides. Textbooks will often give you a routine set of steps. They tell you to subtract the constant on both sides, and then add some constant to both sides, but you can always combine both the subtraction and the addition! If you know that you want a perfect square on the right hand side, then you understand the concept. The objective is not to subtract and add, it is to get a perfect square on the right hand side.

Sometimes, people will also tell you this,

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

That is true, because 4-4=0, but if you understand that we are trying to get a perfect square on both sides, you know that what we really want is

$$y=(x^2+4x+4) - 4+1=(x^2+4x+4)-3$$

which you should be able to see is the same thing as what we got before, just with the 3 on the other side.


Now, you will see why it was important to get a perfect square. We now have

$$y+3 = x^2+4x+4 = (x+2)^2 \implies y=(x+2)^2-3$$

and this is beautiful because we can see that this is exactly the equation for a parabola! If we substitute, for example, $$z=x+2$$, we have $$y=z^2-3$$ and we see that this is just a parabola shifted down three units.

So you see that the reason we wanted a perfect square on the RHS was so we could collapse it into a single quantity squared (in this case x+2). Then, it is very clear that the shape of the graph is a parabola.

Well, that's fine, you could say, but what about my situation, when the equation is

$$y=2x^2-4x+7$$

We again want to make the RHS into some perfect square. We don't want to worry about the 7 right now, so we can, as in your book, move it over to the left hand side, so we have

$$y-7=2x^2-4x$$

It is extremely important that you understand this step is optional! The textbook tells you to do it because it makes looking at the RHS much easier. There isn't a random number there that is distracting you from your objective, which is to make a perfect square.

This is difficult, because we have a 2 in the front. However, we notice that if we divide both sides by 2, then we would have

$$\frac{y-7}{2}=\frac{y}{2}-\frac{7}{2}=x^2+2x$$

Now, we know what to do! Give it a try.

You should strive to understand the big concepts. Going through this method of completing the square, what you should learn is why an equation

$$y=ax^2+bx+c$$

is a parabola. You should also begin to see that every type of equation in that form has an equivalent standard form that you get after completing the square.

Your teacher, your book, and your class will ask you questions such as, find the vertex, etc. These are important, because they show you applications of the standard form, but you should also understand the connection between the form ax^2+bx+c and the standard form a(x-c)^2+d.

 

[kmeyers917]

The problem is the your notion of completing a square. y=2x^2-4x+7 has a perfect square hidden in it. What you need to recognize (and I think you do) is that the 2 in front of the x^2 is hurting you. What I teach is the following
y=2x^2-4x+_+7- _ (the blanks will be filled in later)
y=2(x^2-2x+_)+_
Now that the 2 is outside the () we can complete the square by adding 1 to x^2-2x
(x-1)^2=x^2-2x+1
y=2(x^2-2x+1)+7-2 (notice the 1 in the parenthesis is worth 2 outside the ())
so y =2(x-1)^2+5

In class, I show how the blanks are filled in for instance the x-1 is chosen first because
-1 is half of the coefficient of the middle term(-2x).
The +1 is second to be acquired by squaring the -1.
The -2 next to the 7 is last because it balances out the 2 we added into complete the square.

 

[Mark44]

kmeyers917,
It would be more helpful to provide help in a thread that wasn't almost 2 1/2 years old...