The functions f(x) = –(x + 4)2 + 2 and g(x) = (x − 2)2 − 2 have been rewritten using the completing-the-square method. Is the vertex for each function a minimum or a maximum? Explain your reasoning for each function.

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The functions f(x) = –(x + 4)2 + 2 and g(x) = (x − 2)2 − 2 have been rewritten using the completing-the-square method. Is the vertex for each function a minimum or a maximum? Explain your reasoning for each function.

Mathematics
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If your quadratic has a negative leading coefficient, then it opens down, and its vertex is the aboslute maximum. If your quadratic has a positive leading coefficient, then it opens up, and its vertex is the absolute minimum.
that is the rule.....
Ok so the first one would be that it opens down correct?

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thank-you:) Can I ask anither?
yes, the first one opens down. Right;)
And the second one opens where?
(and yes, you can ask another question right here, after you feel that you are done with this question.)
It opens us since it's poitive
yes
ok i guess you are done with this question... r u?
do you have any questions about this question ?
so the first one, is opening down w/ the vertex at an absolute maximum? I'm just checking
yes
the vertex in function 1 is absolute maximum
Ok thank-you:)
Suppose C and D represent two different school populations where C > D and C and D must be greater than 0. Which of the following expressions is the largest? Explain why. Show all work necessary. (C + D)2 2(C + D) C2 + D2 C2 − D2
yw
i don't understand this one
you don't understand the question, or don't know how to solve it?
I don't understand how to solve it
oh....ok...
I will also add, that C and D must be integers (since you can't have 3.5 people of anything like that).... do you agree with me?
yes. Yes I do
Now, C is at least 1. D is at least 2. (So, C+D is at least 3) Which is the largest? Explain why. Show all work necessary. (C + D)² 2(C + D) C² + D² C² − D²
(C+D)² when C=1 & D=@ is going to be (1+2)²=3²=9 2(C+D) 2(1+2)=2(3)=6
2(C+D) is a linear growth, and (C+D)² is a "quadratic" growth. And the bigger "C+D" we have the greater will (C+D) get when compared to (C+D) that will not grow even nearly as rapidly.
the last option is going to be obviously smaller than or greater than the 3rd option ?
can you tell me?
so the answer would be c2+d2
hold on...
pliz answer my last question...
(you are trying to jump ahead to quickly:o)
smalleer :) Sorry
yes (no need for apologies).
So, the last option falls off... and in competition is the 1st option (C+D)² which is larger than the 2nd option (as i showed), and the 3rd option (which is obviously larger than the 4th option).
((remember D must be greater than C, and both C and D are natural numbers)) C²+D² vs (C+D)² DIFFERENCE ----------------------------------------------------------- C=1 & D=2 1²+2²=5 vs (1+2)²=9 (C+D)² exceeds by 4 ----------------------------------------------------------- C=1 & D=3 1²+3²=10 vs (1+3)²=16 (C+D)² exceeds by 6 ----------------------------------------------------------- C=2 & D=3 2²+3²=13 vs (2+3)²=25 (C+D)² exceeds by 12 ----------------------------------------------------------- C=2 & D=4 2²+4²=18 vs (2+4)²=36 (C+D)² exceeds by 18 (& twice)
The greater values for C and D we pick, the more (C+D)² exceeds, (outmatches if you will) the C²+D². So which one is the largest?
)c+d)^2
Yes, (C+D)² is the largest, and this it is the answer you need:)
Any questions about this problem?
No. :)
OK.....
Three functions are given below: f(x), g(x), and h(x). Explain how to find the axis of symmetry for each function, and rank the functions based on their axis of symmetry (from smallest to largest). f(x) g(x) h(x) f(x) = 3(x + 4)2 + 1 g(x) = 2x2 − 16x + 15
ok, this is a last question for this post. Alright>
Yea:)
h(x) = ?
h(x): graph parabola (1,-3) (3,5) (-1,5)
for h(x), if you were to graph the points:|dw:1435801328063:dw|
you can tell that (1,-3) is the vertex od the parabola.
ys it is the vertex
And that way we know that \(\Large y=\color{red}{\rm a}(x-1)^2-3\) part.
we know everything besides the scale factor (or the coefficient) we will use.
(btw, if you don't undertsnad something completely, you are welcome to ask anything that you would like)
Ok. I understand everything so far. No worries:)
Ok, yes... so now we need to use one of your point. We know that this parabola should satisfy the point (3,5)
don't forget that this parabola that we are finding is the h(x)
So, \(\large\color{black}{ \displaystyle h(x)=\color{red}{\rm a}(x-1)^2-3 }\) we know that point (3,5) should be on the parabola. So: \(\large\color{black}{ \displaystyle \color{blue}{5}=\color{red}{\rm a}(\color{blue}{3}-1)^2-3 }\)
Can you solve for a, please?
ok hold on
a=1/5
5=a(4)-3 5=a1 a=1/5
hello?
@SolomonZelman is it g(x), f(x), and then h(x) ? I'm sorry
\(\large\color{black}{ \displaystyle \color{blue}{5}=\color{red}{\rm a}(\color{blue}{3}-1)^2-3 }\) \(\large\color{black}{ \displaystyle \color{blue}{5}=\color{red}{\rm a}(2)^2-3 }\) \(\large\color{black}{ \displaystyle \color{blue}{5}=4\color{red}{\rm a}-3 }\) \(\large\color{black}{ \displaystyle 2=4\color{red}{\rm a} }\) a=½
this a that we found, we found for h(x).....
So, \(\large\color{black}{ \displaystyle f(x) = 3(x + 4)^2 + 1 }\) \(\large\color{black}{ \displaystyle g(x) = 2x^2 − 16x + 15 }\) \(\large\color{black}{ \displaystyle h(x) = \frac{1}{2}(x -1)^2 -3 }\) (we foudn a=1/2)
I see, I see
and x-axis of summetry (in this case) is just the x-coordinate of the vertex.
just "axis of summetry" (not "x-axis of summetry")
you need to complete the square for g(x)
So, h(x) axis of symmetry would be 1?
@SolomonZelman I have no idea how to do that
yes for h(x) it is 1
can you identify the axis of summetry for f(x) ?
-4
yes
Now, for g(x). \(\large\color{black}{ \displaystyle g(x)=2x^2-16x+15 }\) \(\large\color{black}{ \displaystyle g(x)=2(x^2-8x)+15 }\) following so far?
So i don't get g(x)
we will do g(x)... do you undertsand my previousppost?
previous post*
yes
Ok. \(\large\color{black}{ \displaystyle g(x)=2(x^2-8x)+15 }\) what would you want to have added inside the parenthesis ?
The x's?
what number do you need to add \(\large\color{black}{ \displaystyle g(x)=2(x^2-8x~+\color{blue}{\bf here} )+15 }\) to make the part in the parenthesis a perfect square trinomial ?
2
when you have \(x^2-bx\) the number that you would like to add is \((b/2)^2\)
in this case (8/2)² --> 4² --> 16
So we would like to add 16, but we can't just add numbers, we are going to change the value of the function....
How would we add the number without changing the value of the function? like this....
\(\LARGE \color{black}{ \displaystyle g(x)=2(x^2-8x\color{blue}{+16}\color{red}{-16})+15 }\)
So far I haven't changed anything, I just added a "magic zero"
ok I see
now, expand the -16 (in red) out of parenthesis \(\LARGE \color{black}{ \displaystyle g(x)=2(x^2-8x\color{blue}{+16})+2\cdot (\color{red}{-16})+15 }\) \(\LARGE \color{black}{ \displaystyle g(x)=2(x^2-8x\color{blue}{+16})-32+15 }\) \(\LARGE \color{black}{ \displaystyle g(x)=2(x^2-8x\color{blue}{+16})-17 }\) \(\LARGE \color{black}{ \displaystyle g(x)=2(x-4)^2-17 }\)
so what is the axis of summetry for g(x) ?
4
so would be f(x), hx, gx
yes, now you need to list all axis of summetry in order from smallest to greatest.
go ahead...
fx, hx, gx
yes? No? Maybe so?
yes. f , h, g. -4, 1, 4.
Thank you so much:)
yw

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