Question

Need help with converting regular exponential equations into graphing form and stating the vertex.

Answer 1

\[y=x^2+6x+11\]

Answer 2

Well, first off, be careful - that is NOT an "exponential equation", which means an equation where the variable is in the exponent (e.g., y=3^x) (and they don't have a vertex). :) Just want to get the terminology right.

This is a "quadratic equation". I assume by "graphing form" you are looking for something of the form:

\(y=a(x-h)^2+k\) so that you have the vertex, (h,k), right?

Answer 3

Yeah, sorry. Been awhile since I last had math.

Answer 4

So to do that here, you're going to need to complete the square. Do you know how to do that?

Answer 5

I remember slightly, so I need to factor it and then put both factors on each sides?

Answer 6

No, not quite.... you want to right the equation so that you have a square of a binomial, plus a constant. See, if you look at it as given, you can't factor that polynomial into a square of a binomial (in fact, it doesn't factor at all).

So here's the process: you want to scoot over the constant term.... because we are going to ADD something there:

\(y=x^2+6x+..........+11\)

Answer 7

Now, what are we doing to add? well, we're going to add the constant that is necessary to make the first two terms, \(x^2+6x\), into the first 2 terms of a perfect square trinomial.

What will do that? Take HALF of b (the coefficient on the middle term) and then SQUARE it. Can you tell me what that is, here?

Answer 8

So It'd be 9?

Answer 9

RIGHT... ok, so now we have:

\(y=x^2+6x+9+11\)

Now of course, the problem here is that we just CHANGED THE EQUATION. It isn't the same equation now, so we have to FIX that. How do we fix that? well, we added 9, so we'll just subtract 9 at the other end of the equation:

\(y=x^2+6x+9+11-9\)

Answer 10

do you see how that way, we still have the SAME equation, just with the terms written differently?

Answer 11

If were subtracting 9 again, isn't it like we never added it?

Answer 12

Now you're thinking "ok, great... what was the point in that?" :) Well, now group the first 3 terms together... like so... this will help you better understand what we're up to:

\(y=(x^2+6x+9)+11-9\)

Answer 13

Ohh,

Answer 14

Now look at those 3 terms in the ( )... that is now a special kind of trinomial, called a "perfect square trinomial". That means it factors into the square of a binomial. Do you know how to fact that? (JUST the part in the ( )

Answer 15

*factor that

Answer 16

If by fact you mean get (x+3)^2

Answer 17

YES, exactly! so now I can re-write the equation as:

\(y=(x+3)^2+11-9\)

And then, of course I'm going to add those constant terms to simplify:

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

Answer 18

So H would be -3 while K is 2?

Answer 19

And NOW my equation - the very same equation we stared with - is written differently, in what's usually called "vertex form"
\(y=a(x-h)^2+k\) where (h,k) is the vertex

Answer 20

Yes, exactly - you got it! And you didn't even fall for the "sign trap" on h, good job! :)

Answer 21

Thank you, I never learned it this way, does this work for most equations?

Answer 22

It will certainly work to complete the square. Keep in mind, this one was pretty easy because you had b=6. Easy to half and then square. :)

If you had, say, b=5, the process is the same. (b/2)=5/2, and then the square of that is 25/4. So you add that, and subtract that. A little more messy, but the process is exactly the same.

Answer 23

If you have done circle equations yet, you may have also seen completing the square, there. Or you will if you are going to do them later.

Answer 24

I did, just a little back in my memory. So This next equation I have is where b=7 So it would be 7/2= 49/4?

Answer 25

yes, exactly! :) just means you will have fractions for h and k (some quadratics are just like that) :)

Answer 26

Ok, thanks again!

Answer 27

You're welcome! Happy to help. :)

Answer 28

@DebbieG I'm lost now...