Question

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

Answer 1

Im not really sure how to do this problem

Answer 2

-b/2a

Answer 3

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 \[ax^2+bx+c \]

Answer 4

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

Answer 5

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

Answer 6

so plus 4y?

Answer 7

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

Answer 8

derivatives have nothing to do with finding the vertex.

Answer 9

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

Answer 10

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

Answer 11

4y=x^2+2x+9

Answer 12

like that?

Answer 13

yeah let me double check this for a minute.

Answer 14

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

Answer 15

now divide the entire equation by 4

Answer 16

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 :)

Answer 17

@UsukiDoll

Answer 18

And I just reported you for that comment ^

Answer 19

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

Answer 20

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

Answer 21

i am not familiar with it sorry..

Answer 22

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.

Answer 23

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. :)

Answer 24

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.

Answer 25

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

Answer 26

is that is divided by 4?

Answer 27

keep your equation in fraction form. Decimals are messy.

Answer 28

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

Answer 29

okay i see

Answer 30

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

Answer 31

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.

Answer 32

so far so good everyone.

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

The coefficient of "x^2"

Answer 38

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

Answer 39

okay

Answer 40

\[ax^2+bx+c \] <- standard form

\[\frac{ 1 }{ 4 }x^2+\frac{ 1 }{ 2 }x+\frac{ 9 }{ 4 } \] < - your equation.

Answer 41

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

Answer 42

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.

Answer 43

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

\[\LARGE \frac{-\frac{1}{2}}{2(\frac{1}{4})} \]
\[\LARGE \frac{-\frac{1}{2}}{(\frac{2}{4})} \]

Answer 44

i got (2,1)

Answer 45

wait hold on

Answer 46

(1,?) i almost got it

Answer 47

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 \[\frac{-1}{2} \times \frac{2}{1} \]

Answer 48

(1,2)???

Answer 49

is that right?

Answer 50

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

Answer 51

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

Answer 52

-2 and 2

Answer 53

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

Answer 54

-1

Answer 55

yes that's the first part of our vertex.

Answer 56

(-1,?)

Answer 57

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

Answer 58

2

Answer 59

(-1,2)

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

1/4

Answer 64

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

Answer 65

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

Answer 66

its 2.25

Answer 67

then minus the .25

Answer 68

it equals 2

Answer 69

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

Answer 70

decimals make more sense to me

Answer 71

(-1,2)

Answer 72

fractions are easier to deal with though...

Answer 73

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

Answer 74

yay we got our vertex!

Answer 75

thank you!