anonymous
  • anonymous
Need help with converting regular exponential equations into graphing form and stating the vertex.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[y=x^2+6x+11\]
DebbieG
  • DebbieG
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?
anonymous
  • anonymous
Yeah, sorry. Been awhile since I last had math.

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DebbieG
  • DebbieG
So to do that here, you're going to need to complete the square. Do you know how to do that?
anonymous
  • anonymous
I remember slightly, so I need to factor it and then put both factors on each sides?
DebbieG
  • DebbieG
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: \(y=x^2+6x+..........+11\)
DebbieG
  • DebbieG
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?
anonymous
  • anonymous
So It'd be 9?
DebbieG
  • DebbieG
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: \(y=x^2+6x+9+11-9\)
DebbieG
  • DebbieG
do you see how that way, we still have the SAME equation, just with the terms written differently?
anonymous
  • anonymous
If were subtracting 9 again, isn't it like we never added it?
DebbieG
  • DebbieG
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: \(y=(x^2+6x+9)+11-9\)
anonymous
  • anonymous
Ohh,
DebbieG
  • DebbieG
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 ( )
DebbieG
  • DebbieG
*factor that
anonymous
  • anonymous
If by fact you mean get (x+3)^2
DebbieG
  • DebbieG
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: \(y=(x+3)^2+2\)
anonymous
  • anonymous
So H would be -3 while K is 2?
DebbieG
  • DebbieG
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
DebbieG
  • DebbieG
Yes, exactly - you got it! And you didn't even fall for the "sign trap" on h, good job! :)
anonymous
  • anonymous
Thank you, I never learned it this way, does this work for most equations?
DebbieG
  • DebbieG
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.
DebbieG
  • DebbieG
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.
anonymous
  • anonymous
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?
DebbieG
  • DebbieG
yes, exactly! :) just means you will have fractions for h and k (some quadratics are just like that) :)
anonymous
  • anonymous
Ok, thanks again!
DebbieG
  • DebbieG
You're welcome! Happy to help. :)
anonymous
  • anonymous
@DebbieG |dw:1378155396829:dw| I'm lost now...
DebbieG
  • DebbieG
Ahhhhh.... 2 issues. First of all, this one has a=2, not a=1 like the first one we did, did. You can't get a perfect square trinomial with that leading coefficient of 2. So step 1 is to factor that out: \(2x^2-7x+12=2(x^2-\dfrac{7}{2}) +....... +12\) I went ahead and scooted the 12 over.... NOW figure out your \(b\cdot\dfrac{1}{2}\) and then square it. THEN you must ADD that inside the ( ), because a perfect square trinomial ALWAYS has the last term ADDED. and SUBTRACT it over on the other side of the 12.

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