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What is the vertex for the function? (You will want to find the axis of symmetry first.) Write as an ordered pair, leaving no spaces, such as (1,-4). y = -3x2 + 6x

Mathematics
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im so dumb omg
use the square to convert to a vertex form ,y=a(x-h)^2+k
I guess we're not making sense, @UmarsBack

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Other answers:

you really arent :c idk what im suppsoed to do.
its like : ax^2+bx+c the vertex is on the axis of symmetry: -b/2a when x= -b/2a we find y in this case -3x^2 +6x +0 gives us: a=-3 b=6 so when x=1; we get y= -3+6
So what i gotta do? sorry for my stupidity.
okay let us do some reading ..
http://finedrafts.com/files/math/precal/Larson%20PreCal%208th/Larson%20Precal%20CH2.pdf
x=-b/2a in this case -3x^2 +6x +0 gives us: a=-3 b=6 Which mean x=-6/-6= +1 ! y=-3+6= 3 Vertex = (1,3) =============
Im completely clueless guys.
can you read pp 126-131?
I just want you to be acquainted with the terms and definitions
Yes im reading right now. again, sorry for my stupidity.
@nincompoop im doneee
so what did you learn?
eyad's information should make sense by now...
Quadratic function = f(x) = ax^2 + bx + c
yes, that is the format of your equation at the moment. now, to help you identify the vertex of the parabola (graph of quadratic) you need to covert it into standard form of quadratic function
your vertex then is going to be the h and k
there are a few things you need to be good at like completing the square and factoring so you can solve this even mentally
do you know how to complete the square yet?
wheres the h and k? and no :c
hehe this is going to be tough… so even eyad's info doesn't make sense right now?
Im tryna understand >_<
should i go to khan academy?
you can, but it's in the textbook and eyad pretty much gave the answer already
you know i hate reading lmfao. i read the text book tho. im brain dead right now.
okay let us try this the long way so you understand the whole concept quadratic function f(x) = ax^2 + bx + c; f(x) is the same as "y" so this is what you have: y = -3x2 + 6x; this is the same as y = -3x2 + 6x + 0 (the equation you were given has + 0 omitted) this means that: a = -3 b = 6 c = 0 I need to know that you read and fully grasp this part
Omfg i understand it way more now.
good, now recall the figures in the text book about vertex? I attached it so you can recall them
Yea i recall them
good! we are making a progress here. keep in mind about vertex being either the lowest point or the highest point lowest point if a>0, which means the parabola opens upward highest point if a<0, which means the parabola opens downward the reason I want you to keep this in mind because the terms minimum and maximum give you the same definition in terms of the value of a recall your equation:y = -3x2 + 6x does this mean that your parabola opens upward or downward?
The parabola opens upwards?
I want you to be certain of your answer. Only when you are confident that we are able to move on…
I made the figure bigger so you won't miss any detail
y=-3x^2+6x y=ax^2+bx ^
Its downward. right?
with your equation y = -3x2 + 6x I mentioned this: a = -3 b = 6 c = 0 and I also mentioned this: lowest point if a>0, which means the parabola opens upward highest point if a<0, which means the parabola opens downward ask yourself this question: is the value of a less than or greater than zero? greater than zero gives you an upward parabola less than zero gives you a downward parabola
Its upward, because the value is greater than 0
Btw @nincompoop ,After you finish you may Compile all your comments and post it in a one post ,IT would be a good tutorial Including the example .
awesome! now you seem more confident. now, let us move on understanding the standard form of a quadratic function I attached some information for you to re-read (since you said you read it earlier)
keep in mind that the values of your a, b, and c have not changed. they are still a = -3 b = 6 c = 0
Okay im still with you.
Okay continue. youre the teacher here.
...
now, we need to try to re-arrange your y = -3x^2 + 6x into the standard form of quadratic function, we can achieve it by completing the square, which for now will look like: factor 3 out of x-terms \[y=−3(x^2-2x)+0\] I included the value of c for the purpose of showing you what it would look like, but since our c is zero we can ignore it. do you know how to proceed or know how to complete the square?
I dont know how to complete the square :c
it is now the time to learn one way is to divide the value of b by 2, and then square it but things could get more complicated than that so you will need to read and familiarize yourself http://www.mathsisfun.com/algebra/completing-square.html or this http://www.purplemath.com/modules/sqrquad.htm
6/2= 3 2(3) ? like that?
yes, remember that you have to square it after you divide it by 2
so \[\left( \frac{ b }{ 2 } \right)^2\] \[\left( \frac{ 6 }{ 2 } \right)^2=3^2=?\]
It equals the same thing right? 6?
\[3^2=3\times3=?\] to square is to multiply the multiplicand by itself twice
so your multiplier is also 3
ohh 9.
did you read the contents of the links regarding completing the square I provided earlier?
Not yet. let me read it now.
okay are you done?
Yes, i think.
x2 + bx + (b/2)2 = (x+b/2)2 < thats how you complete the square?
so this is where we left off: \[y=−3(x^2−2x)+0\] we need to complete the square of what is in the parenthesis our b is -2 \[\left( \frac{ -2 }{ 2 } \right)^2=-1^2=?\]
WAKE UP!!! 1 multiplied by itself is 1 this is what we will yield: \[y=-3(x^2-2x+1-1)\] After adding and subtracting 1 within the parentheses, you must now regroup the terms to form a perfect square trinomial. The -1 can be removed from inside the parentheses; however, because of the -3 outside of the parentheses, you must multiply -1 by -3 this is what it would look like: \[y=-3(x^2-2x+1)-3(-1)\] -3(-1) = 3 if you are still with me, we just solved the value of k this confirms one of @eyad's solution earlier k being 3 next step is to factor what is in the parenthesis \[(x^2-2x+1) \] turns into: \[(x-1)^2\]
can you guess the value of h?
1
that is super awesome! our re-written quadratic function form looks like this: \[y=-3(x^2-1)^2+3\] \[y=a(x^2-h)^2+k, where: a \neq0\] vertex: (h,k) (-(-1), 3) or (1,3)
does this look like eyad's answer?
Hahaha wo i cant believe i understand this now. it looks like eyads answer. thanks alot nin omfg.
Yw :)
btw a few notes to consider: \[-1^2 = -(1)^2=-1\] earlier, I should have typed \[(-1)^2=1\] \[-1\times-1=1\] negative multiplied by a negative yields a positive
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