anonymous
  • anonymous
The vertex of x2 + 2x - 4y + 9 = 0 A. (-1,2) B. (1,2) C. (-1,-3) D. (1,-3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Im not really sure how to do this problem
anonymous
  • anonymous
-b/2a
UsukiDoll
  • 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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anonymous
  • anonymous
https://www.khanacademy.org/math/algebra/quadratics/solving_graphing_quadratics/v/finding-the-vertex-of-a-parabola-example
anonymous
  • 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
  • anonymous
so plus 4y?
anonymous
  • anonymous
The to get the vertex we should find the derivative of y and equate it to 0 correct?
UsukiDoll
  • UsukiDoll
derivatives have nothing to do with finding the vertex.
UsukiDoll
  • UsukiDoll
derivatives is in all levels of Calculus... we're just dealing with quadratic equations... which is College Algebra or even lower.
UsukiDoll
  • UsukiDoll
get this equation in this form \[y=ax^2+bx+c \]
anonymous
  • anonymous
4y=x^2+2x+9
anonymous
  • anonymous
like that?
UsukiDoll
  • UsukiDoll
yeah let me double check this for a minute.
UsukiDoll
  • 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
  • UsukiDoll
now divide the entire equation by 4
anonymous
  • 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
  • anonymous
@UsukiDoll
UsukiDoll
  • UsukiDoll
And I just reported you for that comment ^
anonymous
  • anonymous
Dylan are you familiar with derivates or not? So ill continue explaining?
UsukiDoll
  • UsukiDoll
and if OP isn't familiar with derivatives, that technique shouldn't be used.
anonymous
  • anonymous
i am not familiar with it sorry..
UsukiDoll
  • 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
  • anonymous
Hahhahahah 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
  • 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
  • anonymous
y=(x^2/4)+.5+2.25
anonymous
  • anonymous
is that is divided by 4?
UsukiDoll
  • UsukiDoll
keep your equation in fraction form. Decimals are messy.
UsukiDoll
  • UsukiDoll
\[x^2+2x+9=4y \rightarrow \frac{x^2}{4}+\frac{1}{2}x+\frac{9}{4}\]
anonymous
  • anonymous
okay i see
UsukiDoll
  • UsukiDoll
\[\frac{1}{4}x^2+\frac{1}{2}x+\frac{9}{4} = y\]
Owlcoffee
  • 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
  • UsukiDoll
so far so good everyone.
UsukiDoll
  • 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
  • anonymous
how can i plug it in if i don't know what a is?
UsukiDoll
  • 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
  • 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
  • Owlcoffee
The coefficient of "x^2"
UsukiDoll
  • 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
  • anonymous
okay
UsukiDoll
  • UsukiDoll
\[ax^2+bx+c \] <- standard form \[\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 } \] < - your equation.
UsukiDoll
  • UsukiDoll
so a = 1/4, b =1/2, and c=9/4
UsukiDoll
  • 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
  • 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
  • anonymous
i got (2,1)
anonymous
  • anonymous
wait hold on
anonymous
  • anonymous
(1,?) i almost got it
UsukiDoll
  • 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
  • anonymous
(1,2)???
anonymous
  • anonymous
is that right?
anonymous
  • 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
  • UsukiDoll
what's \[\frac{-1}{2} \times \frac{2}{1} \] what's -1 x 2 and what's 2 x 1?
anonymous
  • anonymous
-2 and 2
UsukiDoll
  • UsukiDoll
yes so we have \[\frac{-2}{2} \] now what's -2 divided by 2
anonymous
  • anonymous
-1
UsukiDoll
  • UsukiDoll
yes that's the first part of our vertex.
anonymous
  • anonymous
(-1,?)
UsukiDoll
  • UsukiDoll
\[V(-1,\frac{ 4ac-b^2 }{ 4a }) \] a=1/4, b =1/2, c =9/4
anonymous
  • anonymous
2
anonymous
  • anonymous
(-1,2)
UsukiDoll
  • 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
  • UsukiDoll
\[4ac-b^2 \] is all we need to deal with \[\[4\frac{1}{4}\frac{9}{4}-(\frac{1}{2})^2 \] \]
UsukiDoll
  • 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
  • anonymous
1/4
UsukiDoll
  • UsukiDoll
yeah. \[\[4\frac{1}{4}\frac{9}{4}-\frac{1}{4}\]
UsukiDoll
  • UsukiDoll
so now what's \[4 \times \frac{1}{4} \times \frac{9}{4}\] the fraction parts would be easier to do first
anonymous
  • anonymous
its 2.25
anonymous
  • anonymous
then minus the .25
anonymous
  • anonymous
it equals 2
UsukiDoll
  • UsukiDoll
errrrrrrrr please keep your answers in fraction form.. decimals are messy!
anonymous
  • anonymous
decimals make more sense to me
anonymous
  • anonymous
(-1,2)
UsukiDoll
  • UsukiDoll
fractions are easier to deal with though...
UsukiDoll
  • 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
  • UsukiDoll
yay we got our vertex!
anonymous
  • anonymous
thank you!

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