korosh23
  • korosh23
Pre-cal 12 Question! Writing a polynomial equation. Determine the equation of the polynomial of the least degree that is symmetric to the y-axis, tangent to the x-axis at (3,0) and has P(0)=27.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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korosh23
  • korosh23
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zepdrix
  • zepdrix
Hmm ya! A quadratic should work! :) We're given that: P(0)=27 P(3)=0 (This point must also be the vertex if we're going to try and make a quadratic work) So my thinking is... Our vertex is at (3,0), let's write a generic quadratic in vertex form, and then use the other piece of information to solve for a, the amplitude!
korosh23
  • korosh23
ok

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zepdrix
  • zepdrix
\[\large\rm P(x)=a(x-h)^2+k\]With vertex: \(\large\rm (h,k)\) So our equation thus far should be,\[\large\rm P(x)=a(x-3)^2+0\]Something like that, ya?
korosh23
  • korosh23
exactly
korosh23
  • korosh23
so in yesterday lesson, I learne to find the value of a. making x=0 and plugging in the value of y for p(x), but the ansewr would be a=0
korosh23
  • korosh23
Confusing
zepdrix
  • zepdrix
Your coordinate point is like this:\[\large\rm (x,P(x))\quad= (0,27)\]So we replace x's with 0's, and P(x) with 27, ya?\[\large\rm P(x)=a(x-3)^2\]\[\large\rm 27=a(0-3)^2\]
zepdrix
  • zepdrix
We're calling it P(x), but it's really just y, hopefully you can get comfortable with the function notation :)
korosh23
  • korosh23
yea value of a would be 3 a=3 Thus, P(x)= 3(x-3)^2 + 0
zepdrix
  • zepdrix
Yay good job \c:/ And depending on whether not they want that in `standard form`, you might want to expand out the square. It doesn't look like the instructions were asking for that though.
korosh23
  • korosh23
@zepdrix sure, I still have one question
zepdrix
  • zepdrix
k
zepdrix
  • zepdrix
Hmm hold on, did we screw something up... "symmetric about the y-axis.. thinking
korosh23
  • korosh23
I do not understand what it means
korosh23
  • korosh23
I guess it means both sides should be equal, or look the same
zepdrix
  • zepdrix
symmetric about the y-axis means that it can be reflected across the y-axis and still be the same shape.|dw:1447534382038:dw|Yes, like this example. When reflected over, it covers itself up.
zepdrix
  • zepdrix
|dw:1447534458816:dw|And I guess we DO NOT have that happening with our quadratic huh? :)
korosh23
  • korosh23
Exactly, this leads us to a clue, quadratic function is not possible
zepdrix
  • zepdrix
Hmm interesting :)
zepdrix
  • zepdrix
So we'll actually need something that looks more like this:|dw:1447534509298:dw|
korosh23
  • korosh23
Quintic function, but it did not say x= -3, so why do we use it?
zepdrix
  • zepdrix
If it's symmetric about the y-axis, then P(-3) must be equal to P(3). So if we have P(3)=0, we must also have P(-3)=0
zepdrix
  • zepdrix
Lemme fix my drawing a little bit, because we have P(0)=27.
zepdrix
  • zepdrix
|dw:1447534650930:dw|
korosh23
  • korosh23
Oh I see :D
zepdrix
  • zepdrix
So we have to figure out an equation for this guy? :d Hmm
zepdrix
  • zepdrix
Degree? :)
korosh23
  • korosh23
yes I am gonna give it a try
korosh23
  • korosh23
P(x)= 3(x-3)^2(x+3)^2+27 P(x)= 3(x-9)^4 + 27
zepdrix
  • zepdrix
Hmm not quite. Since we're dealing a 4th degree polynomial now, we're not just jumping to "vertex form" anymore. So it looks like you recognized that the function has `quadratic behavior` at -3 and 3, so in other words, multiplicity of 2 at those roots, ya?\[\large\rm P(x)=a(x-3)^2(x+3)^2\]I think that +27 is out of place, hmm.
zepdrix
  • zepdrix
I wasn't sure where the +27 was coming from, I thought maybe you were doing a vertex thing or something :P
korosh23
  • korosh23
@zepdrix that is my qusetion. Why we do not include y-int in these polynomial equations?
zepdrix
  • zepdrix
Because the y-intercept is formed by multiplying out the brackets. You can't just artificially attach it to the end of the equation.
korosh23
  • korosh23
I get it after multiplying it it shows up. :)
zepdrix
  • zepdrix
What do you get for an a value in this new equation? :)
korosh23
  • korosh23
a= 1/3
zepdrix
  • zepdrix
\[\large\rm P(x)=\frac{1}{3}(x-3)^2(x+3)^2\]Ok great. And if you want to write it in that fancy way you were doing before, applying difference of squares, careful the way you do it,\[\large\rm P(x)=\frac{1}{3}(x-3)(x-3)(x+3)(x+3)\]\[\large\rm P(x)=\frac{1}{3}(x-3)(x+3)(x-3)(x+3)\]\[\large\rm P(x)=\frac{1}{3}(x^2-9)(x^2-9)\]
korosh23
  • korosh23
Perfect, thank you I learned many new things. You are awesome tutor. I got to go for lunch now, my parents are calling me, you know Asian parents lol
zepdrix
  • zepdrix
hehe ^^ have a nice day!
korosh23
  • korosh23
you too my frined

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