Question

what is the standard form of y=0.05x^2+x+1

Answer 1

I was looking at that, and my first reaction, when I learned all this stuff 40 years ago, was that the general and standard forms are the same thing. Looking at the Internet, some authors (not all) are calling the vertex form the standard form. Others are calling the vertex form the general form (few). Some call the standard the general form. Ain't that confusing?

Answer 2

So, the logical thing to do, if you want to get the answer right, from the understanding of your teacher, is to go with what HE says the two forms are.

Answer 3

If you don't know what he says about that, then the consensus of the Internet, from what I can tell, is that the standard form is ax^2 + bx + c = y and the vertex form is the general form. Certainly not the way I learned it.

Answer 4

So, what you have in your question is already the standard form.

Answer 5

it is ....really

Answer 6

hahah that is the standard form sweetie unless u talkin bout graphin it

Answer 7

Yes. And the general form is the vertex form. Terminology should not change like this. So, the general and vertex form is a(x - h)^2 + k = y

Answer 8

no its already graphed i know its a parabola

Answer 9

BTW the "a" in both forms is the same numerical value.

Answer 10

wait so general form is the same as standard....

Answer 11

No, I'm saying that that is the way I learned it and some (a few) authors are still saying that. They must be old like me! :-) The standard is ax^2 + bx + c = y. The general and vertex forms are the same and is a(x - h)^2 + k = y

Answer 12

So, in the vertex form, the actual vertex is (h, k) and the axis of symmetry is x = h.

Answer 13

So, how about we convert your equation to vertex form. Would you like that?

Answer 14

yea a lot actually i would really appreciate it

Answer 15

okay, but I'm going to do it step-by-step so you learn something and not just get an answer. Your promise is to go over everything I did here once you take your test.

Answer 16

ok sounds good

Answer 17

y=0.05x^2+x+1 -> y = 0.05(x^2 + 20x) + 1 ->

y = 0.05(x^2 + 20x + 100) + 1 - 5 -> y = 0.05(x + 10)^2 - 4

I left out a couple of steps, but feel free to ask or at least look at this later so you'll know how to do this in the future.

Answer 18

That second step "20" is from 0.05 x 20 = 1

The thrid step "100" is half of 20, then squared, which is actually 100 x 0.05 or 5, which explains the "-5". Adding and subtracting 5.

Answer 19

I'm throwing a lot at you all at once, so it will be very hard to absorb all this at once if this is all new to you. So, don't despair and get math anxiety. If this is new. Give yourself time to get this. You have to over this again and again and again. Then you will get this. So, stick with it.

Answer 20

okay

Answer 21

It's hard at first, and somewhere along the line, if you do stick with it, and that's the KEY, you will wondeer what the big deal was all along! :-) Really. Treat math like a friend who speaks a foreign language. A good friend.

Answer 22

So, I hope this is helping.

Answer 23

okay thanks a lot :D

Answer 24

Anything more about this equation you need help with?

Answer 25

Can you get the vertex? If you're not sure, that's ok, make a stab.

Answer 26

would the vertx be -4

Answer 27

*vertex

Answer 28

Rember, the vertex is a point, so it will have (x, y) form. It's actually (-h, k) from the vertex form of the equation. So, it would be (-10, -4)

Answer 29

Got a little present for you. See attachment.

Answer 30

There you can see the vertex at (-10, -4). You can also see 2 real zeros and the y-intercept.

Answer 31

on my assignment its a different graph hold on i will post my assignment

Answer 32

you see its different....what do i do??

Answer 33

That's not the real graph. That's like an "advertisement" graph of parabolas in general. It looks like the same shape, but it has to be shifted over to the left quite a bit. Go with my graph.

Answer 34

ok

Answer 35

then how do you get like axis of sem??

Answer 36

Did you generate that graph? Where did it come from? I still stand by my answer like a rock, but I'm curious where that graph came from.

Answer 37

The axis of symmetry is x = -h, so the axis is x = -10. And where did that graph come from?

Answer 38

good question it was on the assigment when it was provided to me....idk

Answer 39

I also saw that equation on your attachment. But they don't go together. Not at all. Either the equation was typo'd or the graph was put in wrong. One or the other. I have 2 confirmations that back up my claim. One is the work I did, and the other is an online graphing tool I used. Both gave the same result. The very first thing to do is verify the equation. If the equation is right, their graph is wrong. If the grpah is right, then we have been working with the wrong equation all along.

Answer 40

well its their fault not mine so i cant get blamed for it
and just asking how or what would be the domain how do you get that?

Answer 41

The domain is the allowable "x" values. You can tell by the end behavior (what is going on at the right and left sides), that you can put in any value for "x". So, for this and all parabolas where "x" is the independent variable, the domain is all x. As for the range, you can see that there is a minimum at y= -4, so the range is y >= -4

Answer 42

I have one more proof that the homework handout graph is wrong. Hold on.

Answer 43

ok....can you actually help me with one more assignment??

Answer 44

\[x = \frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a } = \frac{ -1\pm \sqrt{1^{2}-(4)(0.05)(1)} }{ (2)(0.05) }\]

Answer 45

That gives you the zeros in my graph.

Sure, just start another thread. Don't put it in this one.

Answer 46

That's not the ambiguity. It's about what is termed the "general" form. It's in various places on the Internet. Various authors are conflicting on the term "general", not standard and vertex.