Find the vertex of the graph of the function.
f(x) = 4x^2 + 24x + 32

- anonymous

Find the vertex of the graph of the function.
f(x) = 4x^2 + 24x + 32

- chestercat

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- Jhannybean

Basically you're taking your equation \(y=4x^2+24+32\) that has the quadratic form \(y=ax^2+bx+c\) and converting it into vertex form: \(y=a(x-h)^2+k\)

- Jhannybean

So to start, we can immediately see that all the coefficients are divisible by 4. So simplify your function.

- anonymous

y=4x2+24+32= y=ax2+bx+c
? = y=a(x−h)2+k
right?

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## More answers

- anonymous

a=4
b=24
c=32

- anonymous

y=a(x−h)2+k = = y=4(x−h)2+k

- Jhannybean

You got the idea somewhat, but first lets simplify this :) You'll see what I mean by "fitting the equation into the form" :D

- anonymous

yeah, kinda. what i don't get is how to get x&y values to then find h&k

- anonymous

ohh and thank you so much for helping me @Jhannybean

- Jhannybean

Simplify this: \[\large y=\frac{4}{4}x^2+\frac{24}{4}x+\frac{32}{4}\]

- Jhannybean

- anonymous

thank you

- anonymous

sorry mi wifi was acting up @Jhannybean

- anonymous

y=x2+6x+8

- anonymous

y=x2+6x+8 y=ax2+bx+c
? y=a(x−h)2+k

- anonymous

but how will i get the other values?

- Jhannybean

Alright, now that we simplified it to \(y=x^2+6x+8\), we need to use the method of completing the square. Have you learned that method yet?

- anonymous

not yet

- Jhannybean

In the method of completing the square, we will basically create another quadratic function into the quadratic function we already have, \(y=x^2+6x+8\) It's like an inception of quadratics that will help you find the vertex. :P

- Jhannybean

So let's work it out slowly. First set your function equal to zero. \(y=x^2+6x+8=0\)
Next, move over your constants to one side of the equation, while leaving your variables on the other side side. \(y=x^2+6x=-8\)
Following so far?

- anonymous

yes

- anonymous

do i get x and y intercepts?

- Jhannybean

Once we're dne you'll have your x and y intercepts and you'll see what I mean, just follow me on this one :P Trust!

- Jhannybean

Now we're going to create a quadratic on the left hand side of the function by completing the square of the left hand side. Essentially we're going to `finish` the quadratic on the left hand side so it resembles the form \(ax^2+bx+c\) but right now, we've only got \(ax^2+bx\) and THEREFORE need to find ourselves a new \(c\) value.

- Jhannybean

\[c=\left(\frac{b}{2}\right)^2 \longrightarrow c=\left(\frac{6}{2}\right)^2 = (3)^2 = 9\]

- anonymous

so c = 9

- Jhannybean

Yep. So since we're `adding` c to the left side, we have to respectively add it to the right side as well

- Jhannybean

\[y=x^2+6x\color{red}{+9}=-8\color{red}{+9}\]

- anonymous

y=x2+6x+9=−8+9
y=x2+6x+9=+1

- anonymous

ohh ok

- Jhannybean

If you're good with simplifying quadratics, you'll see that the left hand side now is a PERFECT square! :o \(x^2+6x+9 \longrightarrow (x+3)^2\)

- Jhannybean

Rewrite it. \[y=(x+3)^2=1\]

- Jhannybean

Well, you would move the 1 over to fit your new equation to fit the vertex form \(y=a(x+h)^2+k\)

- Jhannybean

Therefore \(y=a(x+h)^2+k \longrightarrow y=(x+3)^2-1\)

- Jhannybean

Do you see how it works?

- Jhannybean

Whoop, I wrote it wrong, \(y=a(x-h)^2+k\)** Typo.

- anonymous

y=a(x−h)2+k⟶y=(x+3)2−1

- anonymous

⟶y=(x+3)^2−1

- Jhannybean

So we know a positive can be split into 2 negatives, therefore \(y=a(x-h)^2+k \longrightarrow y=(x-(-3))^2-1\)

- Jhannybean

So tell me, what are your x and y intercepts? :D

- anonymous

y=-1

- anonymous

xintercept = (x−(−3))2−1 =
1 = (x−(−3))2

- Jhannybean

Im not understanding that.

- anonymous

1 = (x−(−3))^2

- Jhannybean

\[y=a(x\color{red}{-h})^2\color{blue}{+k} \longrightarrow y=(x-\color{red}{(-3)})^2\color{blue}{-1}\]

- Jhannybean

To help you identify them a little better.
the part in red is your x-intercept, part in blue is your y intercept

- anonymous

thats much better, sorry i was confused

- anonymous

\[y=a(x−h)^{2}+k⟶y=(x−(−3))^{2}−1\]

- anonymous

wait so what do i do now? the intercepts?

- Jhannybean

You're toldto find the vertex of the graph, which I helped you find in the post above :)

- anonymous

thank you @Jhannybean , sorry im a little heard-headed but i just reread it and i completely understand

- Jhannybean

Awesome! Glad I could help :)

- anonymous

##### 1 Attachment

- anonymous

@Jhannybean sorry to bother you again

- anonymous

does it mean that the answer would be "b"?

- anonymous

im so confused :(

- anonymous

use is formula....\[-\frac{ b }{ 2a }\]

- anonymous

do i plug in a=1 and b=6??

- anonymous

\[y=ax^2+bx+c\]

- anonymous

y=(x−(−3))^2−1
-\[\frac{ -3 }{ 1 }\]

- anonymous

-(-3/2)

- anonymous

right?

- anonymous

|dw:1438666779185:dw|

- anonymous

3

- anonymous

a and b is coefficient

- anonymous

all the numberse

- anonymous

24/8

- anonymous

add a negative sign

- anonymous

-3

- anonymous

yea

- anonymous

##### 1 Attachment

- anonymous

- anonymous

would it be the -3,-4
or -4,-3

- anonymous

lowest point is obviously the y-axis

- anonymous

im so lost

- dan815

dantheman

- dan815

lisnten to me, do u know how transformations work with functions

- anonymous

GET A LIFE DAN SRS

- anonymous

no

- anonymous

sorry

- dan815

do you know how transformations work??

- anonymous

kind of

- dan815

okay how about completing the square

- anonymous

f(x-3) = f(x) shifted 3 to the right as example

- anonymous

ok i get that

- anonymous

if i am subtracing 3 away from what i am inputting at the time, then it must appear later by 3 units

- dan815

f(x) = 4x^2 + 24x + 32
the point im trying to get across is
we want to write this equation in the form
f(x) = a(x-b)^2+c
this means the vertex is at (b,c)

- anonymous

i know its - 3

- dan815

okay forget that BS for now

- dan815

do you get why this equation has vertex at b,c though

- anonymous

what i dont know
is if it goes before or after the -4

- dan815

f(x) = a(x-b)^2+c

- dan815

no forget all that stuff, we are going to do this logically

- dan815

f(x) = a(x-b)^2+c
do you get why the vertex for this equation is b,c

- dan815

think about the base graph

- dan815

y=x^2

- dan815

how do you apply transformations to it?

- dan815

y=(x-b)^2
would shift it b units to the right

- dan815

y=(x-b)^2 + c would shift it b units to the right and c units up

- dan815

you can think about it like this too
(y-c)=(x-b)^2
y shifted c up and x shifted b right

- anonymous

THIS MAN HAS CLEARLY LOST IT

- anonymous

sorry my wifi wasn't working

- anonymous

okay
im following

- dan815

do u get why the vertex for
y=a(x-b)^2+c is at (b,c) then?

- anonymous

yes

##### 1 Attachment

- dan815

okay then
y=4x^2 + 24x + 32
rewrite this in that form

- anonymous

y=a(x-b)^2+c
y=4(x-24)^2+32

- dan815

thats not how it works

- dan815

expand the brackets see if u get the right equation

- anonymous

y=x^2+6x+8

- anonymous

ohhhh

- dan815

y=4(x-24)^2+32
=4*(x-24)(x-24)+32
= 4*(x^2-48x+24^2)+32

- dan815

its much bigger than before now

- dan815

so we have to transform to the form we want properly

- anonymous

okay so you did the square

- dan815

okay ill show u how to do it this time, u can try to do it later

- anonymous

(a+b)^2=a^2+2ab+b^2

- anonymous

= 4*(x^2-48x+24^2)+32

- dan815

y=4x^2 + 24x + 32
y=4(x^2+6x)+32
ill complete the square in the brackets
=4(x^2+6x+9-9)+32
=4((x^2+6x+9)-9)+32
=4*((x+3)^2-9)+32
=4*(x+3)^2+32-9*4
=4*(x+3)^2+32-36
=4*(x+3)^2-4
so the vertex is at
(-3,-4)

- anonymous

thank you so much

- anonymous

uhhhhh finally i understand

- dan815

y=4x^2 + 32x + 32
tell me the vertex for this equation and i will believe you

- dan815

follow the steps i went through up there, its a very similar question

- anonymous

@dan815 whoa you went thru all at

- dan815

it looks like a lot of steps because im showing every little change, but normally once u learn the method u do it all in 1 or 2 steps

- dan815

u can do these in ur head pretty much

- anonymous

yea ikr thats why i said that lol

- anonymous

y=4x^2 + 32x + 32
y=4(x^2+8x)+32
would it be that

- anonymous

so i can continue

- dan815

thats a start now u have to complete the square in those brackets go on

- anonymous

y=4x^2 + 32x + 32
y=4(x^2+8x)+32
=4(x^2+8x+9-9)+32
=4((x^2+8x+9)-9)+32
=4*((x+4)^2-9)+32
=4*(x+4)^2+32-9*4
=4*(x+4)^2+32-36
=4*(x+4)^2-4

- anonymous

:w

- dan815

no

- dan815

the question u are asking urself here is
how can i get a perfect square out of
x^2+8x

- dan815

when u see x^2+8x you have to realize that
the square has to be (x+4)^2 since this is the only square that will give u a x^2 and 8x
now the constant is not there so we add and subtract that the constant part will be 4^2=16 so
x^2+8x+16-16

- anonymous

so you have to multiply 8 by 2

- dan815

u are not multiplying 8 by 2

- dan815

u are dividing (8/2) then squaring
(8/2)^2 = 16
just so happesn that its also the same as 8*2

- anonymous

y=4x^2 + 32x + 32
y=4(x^2+8x)+32
=4(x^2+8x+16-16)+32
=4((x^2+8x+16)-16)+32
=4*((x+4)^2-16)+32
=4*(x+4)^2+32-16*4

- dan815

when u have
x^2+ax
and u want to write it in square form
you take half of a
as
(x+a/2)^2=x^2+ax+(a/2)^2
^--- term on x works out

- dan815

yes now u got it so what is the vertex

- anonymous

-4,-4??

- dan815

no

- dan815

what is the constant term

- anonymous

16

- dan815

=4*(x+4)^2+32-16*4
simplfy this to
a(x-b)^2+c

- dan815

=4*(x+4)^2+32-16*4
=4*(x+4)^2+32-64
=??

- anonymous

-32

- anonymous

=4*(x+4)^2-32

- dan815

oaky so what is the vertex?

- anonymous

i keep thinking its -4,-3

- anonymous

but i don't think it is that

- dan815

what is the vertex for this equation
=a(x-b)^2+c

- anonymous

b,c

- dan815

okay so what is the vertex for this
=4*(x+4)^2-32

- anonymous

-4,-32

- dan815

yes

- anonymous

wow

- dan815

:)

- anonymous

that was really hard but now i got it lol

- anonymous

only the inside changes sign, its the opposite

- anonymous

thank you dan815

- anonymous

you're a lifesaver!

- dan815

no problem, try some completing the square excercises just to make sure though

- dan815

or just do some expanding excercised, that should build your intuition up

- anonymous

i tried doing it with analytic geometry method lol

- anonymous

thank you

- anonymous

works too

- dan815

dantheman

- dan815

for the record here is a short cut

- anonymous

would it be the same for Writing the quadratic function in vertex form.
y = x2 + 8x + 18

- dan815

f(x) = 4x^2 + 24x + 32
you take the middle number and divide by -2* the first number
24/(-2*4) = -24/8=-3
now plug -3 into equation
f(-3)=4*9+24*-3+32=-4
vertex is at (3,f(-3)) = (-3,-4)

- anonymous

ohh wow that as much faster

- dan815

NO do not use it

- dan815

since u do not understand it

- dan815

not until you understand completing the square method, then u can derive this

- dan815

and then u can derive it one more time after u learn about derivatives

- anonymous

yea i told him that stuff earlier lol -b/2a

- dan815

f(x)=ax^2+bx+c
prove that the vertex is at (-b/2a, )
x_coord=-b/2a
y_coord=f(-b/2a)
This can be your home work or something

- anonymous

thank you

- dan815

hint complete square for
f(x)=ax^2+bx+c

- anonymous

:)

- anonymous

how can i rate a qualified helper

- anonymous

(5 starts)

- anonymous

lol

- Jhannybean

Similar to my method without simplifying the equation first

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