Ask Me Help Desk - Converting Qudratic form into Vertex form?

Here's the problem:
f(x)= x^2+18x+72

18/2= 9^2= 81

and then here's the part where I'm confused with:
our teacher tells us:

f(x)= (x^2+18x+81)+72-81= -9 <------- she tells us to add 81 inside of the perentinces (sorry, my right-click is kind of broken at this moment, bear with me on the misspellings) and then to subtract it to the number outside.

vertex form: f(x)= (x+9)^2-9

My math teacher is quite the "cheesy" per say type, and always makes a mistake with her problems, unless a student corrects her. It makes me feel like I can't trust her with teaching me these things, which I need to pass! :confused:

Quote:

Originally Posted by TL4ever

f(x)= (x^2+18x+81)+72-81= -9 <------- she tells us to add 81 inside of the perentinces

Well, the first two terms should have been put inside parenthesis to begin with:

$(x^2\,+\,18x\,\,\,\,\,\,\,\,\,\,\,\,\) + \,72$

Leaving the blank space for your number that "completes the square." You want this part set off from the constant.

We have a rule that says we can't just randomly do something to one side of an equation only, or we won't be equal anymore. However, we need that +81 in there to complete the square. So we have to offset this somehow in order to stay equal. We're not going to add 81 to the left side as well cause that's not what we're attempting to do.

Well, how about instead we subtract another 81 from the right side? So we added one, and now we subtract one, it nets itself out, and we are still equal.

$(x^2\,+\,18x\,+\,81)\, +\, 72\, -\, 81$

Then the part in parenthesis is $(x+9)^2$ as you already have. The rest stays outside the parenthesis. Only the part outside the parenthesis equals -9. That "= -9" is in an inappropriate place cause the whole thing doesn't equal -9.

Quote:

Originally Posted by morgaine300

Well, the first two terms should have been put inside parenthesis to begin with:

$(x^2\,+\,18x\,\,\,\,\,\,\,\,\,\,\,\,\) + \,72$

Leaving the blank space for your number that "completes the square." You want this part set off from the constant.

We have a rule that says we can't just randomly do something to one side of an equation only, or we won't be equal anymore. However, we need that +81 in there to complete the square. So we have to offset this somehow in order to stay equal. We're not going to add 81 to the left side as well cause that's not what we're attempting to do.

Well, how about instead we subtract another 81 from the right side? So we added one, and now we subtract one, it nets itself out, and we are still equal.

$(x^2\,+\,18x\,+\,81)\, +\, 72\, -\, 81$

Then the part in parenthesis is $(x+9)^2$ as you already have. The rest stays outside the parenthesis. Only the part outside the parenthesis equals -9. That "= -9" is in an inappropriate place cause the whole thing doesn't equal -9.

Oh, I see, thanks!