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anonymous
 one year ago
The vertex of x2 + 2x  4y + 9 = 0
A. (1,2)
B. (1,2)
C. (1,3)
D. (1,3)
anonymous
 one year ago
The vertex of x2 + 2x  4y + 9 = 0 A. (1,2) B. (1,2) C. (1,3) D. (1,3)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im not really sure how to do this problem

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2it's one of those shortcut formulas When working with the vertex form of a quadratic equation \[h = \frac{b}{2a}\] and k=f(h) . The a and b portions come from \[ax^2+bx+c \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok look first let us get the y to a side alone so you get: y=x^2/4 + x/2 + 9/2 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The to get the vertex we should find the derivative of y and equate it to 0 correct?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2derivatives have nothing to do with finding the vertex.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2derivatives is in all levels of Calculus... we're just dealing with quadratic equations... which is College Algebra or even lower.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2get this equation in this form \[y=ax^2+bx+c \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yeah let me double check this for a minute.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[x^2 + 2x  4y + 9 = 0 \rightarrow x^2+2x+9=4y \] because we are sort of dealing with this backwards, but I don't think it should be a problem

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2now divide the entire equation by 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh really smartipants let me enlighten you the derivative is the slope of the tan at a specific pt so when you find the derivative of a function and you equate to 0 that means your dealing with the line parallel to the x axis (slope=0) which is the vertex hahahaha everything works on derivates you clever one I dont advise to tell anybody this again because they'll laugh at you :)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2And I just reported you for that comment ^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Dylan are you familiar with derivates or not? So ill continue explaining?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2and if OP isn't familiar with derivatives, that technique shouldn't be used.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am not familiar with it sorry..

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2See I told you that's a bad idea @joyraheb. And you trying to outsmart me makes it even worse. No one should trust anyone with a lower SS.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hahhahahah report thats your way to defend your mathematics 😂 did I even curse well I'll give you an advise, beware of your false knowledge; it is more dangerous than ignorance. :)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Being rude on here counts as getting you reported man... and Dylan already admit to not knowing derivatives, so that technique you are suggesting isn't going to work.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that is divided by 4?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2keep your equation in fraction form. Decimals are messy.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[x^2+2x+9=4y \rightarrow \frac{x^2}{4}+\frac{1}{2}x+\frac{9}{4}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{1}{4}x^2+\frac{1}{2}x+\frac{9}{4} = y\]

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0Any parabola can be represented by the parallelism they posses on either axis of reference (xoy). We can take the expression you posted to the structure: \[y=ax^2+bx+c\] This is also the equation that represents a parabola parallel to the yaxis, so I will, without proof give you the structure of the coordinates that represent the vertice: \[V(\frac{ b }{ 2a },\frac{ 4acb^2 }{ 4a })\] So, taking the conic equation you gave: \[x^2+2x4y+9=0\] We will isolate the "y" term in order to make the coefficients "a", "b" and "c" evident: \[y=\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 }\] And then you use the structure I gave you, to find the coordinates of the vertex.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so far so good everyone.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so now we let \[a= \frac{1}{4}, b =\frac{1}{2}, c =\frac{9}{4} \] and plug it into \[V(\frac{ b }{ 2a },\frac{ 4acb^2 }{ 4a }) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how can i plug it in if i don't know what a is?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2remember our quadratic equation . here is the standard form \[ax^2+bx+c \] and you have this equation \[y=\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 }\] so comparing these two equations we have our a b and c

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but you just said a = the equation and i don't know what a is so i can't plug it in the vertex formula

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.0The coefficient of "x^2"

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2please read my previous comment. you can get a, b, and c, by comparing your equation to the standard. you will see that a = 1/4, b =1/2, and c=9/4

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[ax^2+bx+c \] < standard form \[\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 } \] <  your equation.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so a = 1/4, b =1/2, and c=9/4

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2our vertex formula \[V(\frac{ b }{ 2a },\frac{ 4acb^2 }{ 4a }) \] you just have to plug a, b, and c into these formulas, then we have our vertex.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2let's do the first one together if b = 1/2 and a = 1/4 \[\frac{b}{2a} \] so we place b =1/2 and a =1/4 in that formula \[\LARGE \frac{\frac{1}{2}}{2(\frac{1}{4})} \] \[\LARGE \frac{\frac{1}{2}}{(\frac{2}{4})} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(1,?) i almost got it

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2well let's go a bit slowly on this . \[\LARGE \frac{\frac{1}{2}}{(\frac{2}{4})} \] the bottom fraction can be reduced further if we divide 2 on the numerator and denominator on 2/4 we have 1/2 \[\LARGE \frac{\frac{1}{2}}{(\frac{1}{2})} \] then flip the second fraction and we should have \[\frac{1}{2} \times \frac{2}{1} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i just got confused on plugging it in but then i did it what i think was right and thats the answer i got

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2what's \[\frac{1}{2} \times \frac{2}{1} \] what's 1 x 2 and what's 2 x 1?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yes so we have \[\frac{2}{2} \] now what's 2 divided by 2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yes that's the first part of our vertex.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[V(1,\frac{ 4acb^2 }{ 4a }) \] a=1/4, b =1/2, c =9/4

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2we're lucky because the denominator will just be 1 so we just have the numerator to deal with \[\frac{4}{4} \rightarrow 1 \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[4acb^2 \] is all we need to deal with \[\[4\frac{1}{4}\frac{9}{4}(\frac{1}{2})^2 \] \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so let's deal with the right part of this formula. what's \[(\frac{1}{2})^2 \rightarrow \frac{1}{2} \times \frac{1}{2} \]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yeah. \[\[4\frac{1}{4}\frac{9}{4}\frac{1}{4}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so now what's \[4 \times \frac{1}{4} \times \frac{9}{4}\] the fraction parts would be easier to do first

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2errrrrrrrr please keep your answers in fraction form.. decimals are messy!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0decimals make more sense to me

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2fractions are easier to deal with though...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[4 \times \frac{1}{4} \times \frac{9}{4} \frac{1}{4} \] \[4 \times \frac{9}{16} \frac{1}{4} \] \[\frac{36}{16} \frac{1}{4} \] reducing the left fraction by dividing 4 on the numerator and denominator \[\frac{9}{4} \frac{1}{4} \] same numerator so we have 8/4 which is 2 :)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yay we got our vertex!
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