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

- anonymous

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

Im not really sure how to do this problem

- anonymous

-b/2a

- UsukiDoll

it'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 \[ax^2+bx+c \]

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## More answers

- anonymous

https://www.khanacademy.org/math/algebra/quadratics/solving_graphing_quadratics/v/finding-the-vertex-of-a-parabola-example

- anonymous

Ok look first let us get the y to a side alone so you get: y=x^2/4 + x/2 + 9/2 right?

- anonymous

so plus 4y?

- anonymous

The to get the vertex we should find the derivative of y and equate it to 0 correct?

- UsukiDoll

derivatives have nothing to do with finding the vertex.

- UsukiDoll

derivatives is in all levels of Calculus... we're just dealing with quadratic equations... which is College Algebra or even lower.

- UsukiDoll

get this equation in this form
\[y=ax^2+bx+c \]

- anonymous

4y=x^2+2x+9

- anonymous

like that?

- UsukiDoll

yeah let me double check this for a minute.

- UsukiDoll

\[x^2 + 2x - 4y + 9 = 0 \rightarrow x^2+2x+9=4y \] because we are sort of dealing with this backwards, but I don't think it should be a problem

- UsukiDoll

now divide the entire equation by 4

- anonymous

Oh 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 :)

- anonymous

@UsukiDoll

- UsukiDoll

And I just reported you for that comment ^

- anonymous

Dylan are you familiar with derivates or not? So ill continue explaining?

- UsukiDoll

and if OP isn't familiar with derivatives, that technique shouldn't be used.

- anonymous

i am not familiar with it sorry..

- UsukiDoll

See 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

Hahhahahah report thats your way to defend your mathematics ðŸ˜‚ did I even curse well I'll give you an advise, beware of your false knowledge; it is more dangerous than ignorance. :)

- UsukiDoll

Being 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

y=(x^2/4)+.5+2.25

- anonymous

is that is divided by 4?

- UsukiDoll

keep your equation in fraction form. Decimals are messy.

- UsukiDoll

\[x^2+2x+9=4y \rightarrow \frac{x^2}{4}+\frac{1}{2}x+\frac{9}{4}\]

- anonymous

okay i see

- UsukiDoll

\[\frac{1}{4}x^2+\frac{1}{2}x+\frac{9}{4} = y\]

- Owlcoffee

Any 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 y-axis, so I will, without proof give you the structure of the coordinates that represent the vertice:
\[V(-\frac{ b }{ 2a },\frac{ 4ac-b^2 }{ 4a })\]
So, taking the conic equation you gave:
\[x^2+2x-4y+9=0\]
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

so far so good everyone.

- UsukiDoll

so now we let \[a= \frac{1}{4}, b =\frac{1}{2}, c =\frac{9}{4} \] and plug it into
\[V(-\frac{ b }{ 2a },\frac{ 4ac-b^2 }{ 4a }) \]

- anonymous

how can i plug it in if i don't know what a is?

- UsukiDoll

remember 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

but 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

The coefficient of "x^2"

- UsukiDoll

please 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

- anonymous

okay

- UsukiDoll

\[ax^2+bx+c \] <- standard form
\[\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 } \] < - your equation.

- UsukiDoll

so a = 1/4, b =1/2, and c=9/4

- UsukiDoll

our vertex formula
\[V(-\frac{ b }{ 2a },\frac{ 4ac-b^2 }{ 4a }) \]
you just have to plug a, b, and c into these formulas, then we have our vertex.

- UsukiDoll

let'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

i got (2,1)

- anonymous

wait hold on

- anonymous

(1,?) i almost got it

- UsukiDoll

well 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

(1,2)???

- anonymous

is that right?

- anonymous

i just got confused on plugging it in but then i did it what i think was right and thats the answer i got

- UsukiDoll

what's \[\frac{-1}{2} \times \frac{2}{1} \]
what's -1 x 2
and what's 2 x 1?

- anonymous

-2 and 2

- UsukiDoll

yes so we have \[\frac{-2}{2} \]
now what's -2 divided by 2

- anonymous

-1

- UsukiDoll

yes that's the first part of our vertex.

- anonymous

(-1,?)

- UsukiDoll

\[V(-1,\frac{ 4ac-b^2 }{ 4a }) \]
a=1/4, b =1/2, c =9/4

- anonymous

2

- anonymous

(-1,2)

- UsukiDoll

we're lucky because the denominator will just be 1 so we just have the numerator to deal with \[\frac{4}{4} \rightarrow 1 \]

- UsukiDoll

\[4ac-b^2 \] is all we need to deal with
\[\[4\frac{1}{4}\frac{9}{4}-(\frac{1}{2})^2 \] \]

- UsukiDoll

so let's deal with the right part of this formula. what's \[(\frac{1}{2})^2 \rightarrow \frac{1}{2} \times \frac{1}{2} \]

- anonymous

1/4

- UsukiDoll

yeah. \[\[4\frac{1}{4}\frac{9}{4}-\frac{1}{4}\]

- UsukiDoll

so now what's \[4 \times \frac{1}{4} \times \frac{9}{4}\]
the fraction parts would be easier to do first

- anonymous

its 2.25

- anonymous

then minus the .25

- anonymous

it equals 2

- UsukiDoll

errrrrrrrr please keep your answers in fraction form.. decimals are messy!

- anonymous

decimals make more sense to me

- anonymous

(-1,2)

- UsukiDoll

fractions are easier to deal with though...

- UsukiDoll

\[4 \times \frac{1}{4} \times \frac{9}{4} -\frac{1}{4} \]
\[4 \times \frac{9}{16} -\frac{1}{4} \]
\[\frac{36}{16} -\frac{1}{4} \]
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

yay we got our vertex!

- anonymous

thank you!

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