anonymous
  • anonymous
Find the vertex of the graph of the function. f(x) = 4x^2 + 24x + 32
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Jhannybean
  • 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
  • Jhannybean
So to start, we can immediately see that all the coefficients are divisible by 4. So simplify your function.
anonymous
  • anonymous
y=4x2+24+32= y=ax2+bx+c ? = y=a(x−h)2+k right?

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anonymous
  • anonymous
a=4 b=24 c=32
anonymous
  • anonymous
y=a(x−h)2+k = = y=4(x−h)2+k
Jhannybean
  • 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
  • anonymous
yeah, kinda. what i don't get is how to get x&y values to then find h&k
anonymous
  • anonymous
ohh and thank you so much for helping me @Jhannybean
Jhannybean
  • Jhannybean
Simplify this: \[\large y=\frac{4}{4}x^2+\frac{24}{4}x+\frac{32}{4}\]
Jhannybean
  • Jhannybean
@DanTheMan99
anonymous
  • anonymous
thank you
anonymous
  • anonymous
sorry mi wifi was acting up @Jhannybean
anonymous
  • anonymous
y=x2+6x+8
anonymous
  • anonymous
y=x2+6x+8 y=ax2+bx+c ? y=a(x−h)2+k
anonymous
  • anonymous
but how will i get the other values?
Jhannybean
  • 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
  • anonymous
not yet
Jhannybean
  • 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
  • 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
  • anonymous
yes
anonymous
  • anonymous
do i get x and y intercepts?
Jhannybean
  • 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
  • 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
  • Jhannybean
\[c=\left(\frac{b}{2}\right)^2 \longrightarrow c=\left(\frac{6}{2}\right)^2 = (3)^2 = 9\]
anonymous
  • anonymous
so c = 9
Jhannybean
  • 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
  • Jhannybean
\[y=x^2+6x\color{red}{+9}=-8\color{red}{+9}\]
anonymous
  • anonymous
y=x2+6x+9=−8+9 y=x2+6x+9=+1
anonymous
  • anonymous
ohh ok
Jhannybean
  • 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
  • Jhannybean
Rewrite it. \[y=(x+3)^2=1\]
Jhannybean
  • 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
  • Jhannybean
Therefore \(y=a(x+h)^2+k \longrightarrow y=(x+3)^2-1\)
Jhannybean
  • Jhannybean
Do you see how it works?
Jhannybean
  • Jhannybean
Whoop, I wrote it wrong, \(y=a(x-h)^2+k\)** Typo.
anonymous
  • anonymous
y=a(x−h)2+k⟶y=(x+3)2−1
anonymous
  • anonymous
⟶y=(x+3)^2−1
Jhannybean
  • 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
  • Jhannybean
So tell me, what are your x and y intercepts? :D
anonymous
  • anonymous
y=-1
anonymous
  • anonymous
xintercept = (x−(−3))2−1 = 1 = (x−(−3))2
Jhannybean
  • Jhannybean
Im not understanding that.
anonymous
  • anonymous
1 = (x−(−3))^2
Jhannybean
  • Jhannybean
\[y=a(x\color{red}{-h})^2\color{blue}{+k} \longrightarrow y=(x-\color{red}{(-3)})^2\color{blue}{-1}\]
Jhannybean
  • 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
  • anonymous
thats much better, sorry i was confused
anonymous
  • anonymous
\[y=a(x−h)^{2}+k⟶y=(x−(−3))^{2}−1\]
anonymous
  • anonymous
wait so what do i do now? the intercepts?
Jhannybean
  • Jhannybean
You're toldto find the vertex of the graph, which I helped you find in the post above :)
anonymous
  • anonymous
thank you @Jhannybean , sorry im a little heard-headed but i just reread it and i completely understand
Jhannybean
  • Jhannybean
Awesome! Glad I could help :)
anonymous
  • anonymous
anonymous
  • anonymous
@Jhannybean sorry to bother you again
anonymous
  • anonymous
does it mean that the answer would be "b"?
anonymous
  • anonymous
im so confused :(
anonymous
  • anonymous
use is formula....\[-\frac{ b }{ 2a }\]
anonymous
  • anonymous
do i plug in a=1 and b=6??
anonymous
  • anonymous
\[y=ax^2+bx+c\]
anonymous
  • anonymous
y=(x−(−3))^2−1 -\[\frac{ -3 }{ 1 }\]
anonymous
  • anonymous
-(-3/2)
anonymous
  • anonymous
right?
anonymous
  • anonymous
|dw:1438666779185:dw|
anonymous
  • anonymous
3
anonymous
  • anonymous
a and b is coefficient
anonymous
  • anonymous
all the numberse
anonymous
  • anonymous
24/8
anonymous
  • anonymous
add a negative sign
anonymous
  • anonymous
-3
anonymous
  • anonymous
yea
anonymous
  • anonymous
anonymous
  • anonymous
@saseal
anonymous
  • anonymous
would it be the -3,-4 or -4,-3
anonymous
  • anonymous
lowest point is obviously the y-axis
anonymous
  • anonymous
im so lost
dan815
  • dan815
dantheman
dan815
  • dan815
lisnten to me, do u know how transformations work with functions
anonymous
  • anonymous
GET A LIFE DAN SRS
anonymous
  • anonymous
no
anonymous
  • anonymous
sorry
dan815
  • dan815
do you know how transformations work??
anonymous
  • anonymous
kind of
dan815
  • dan815
okay how about completing the square
anonymous
  • anonymous
f(x-3) = f(x) shifted 3 to the right as example
anonymous
  • anonymous
ok i get that
anonymous
  • anonymous
if i am subtracing 3 away from what i am inputting at the time, then it must appear later by 3 units
dan815
  • 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
  • anonymous
i know its - 3
dan815
  • dan815
okay forget that BS for now
dan815
  • dan815
do you get why this equation has vertex at b,c though
anonymous
  • anonymous
what i dont know is if it goes before or after the -4
dan815
  • dan815
f(x) = a(x-b)^2+c
dan815
  • dan815
no forget all that stuff, we are going to do this logically
dan815
  • dan815
f(x) = a(x-b)^2+c do you get why the vertex for this equation is b,c
dan815
  • dan815
think about the base graph
dan815
  • dan815
y=x^2
dan815
  • dan815
how do you apply transformations to it?
dan815
  • dan815
y=(x-b)^2 would shift it b units to the right
dan815
  • dan815
y=(x-b)^2 + c would shift it b units to the right and c units up
dan815
  • dan815
you can think about it like this too (y-c)=(x-b)^2 y shifted c up and x shifted b right
anonymous
  • anonymous
THIS MAN HAS CLEARLY LOST IT
anonymous
  • anonymous
sorry my wifi wasn't working
anonymous
  • anonymous
okay im following
dan815
  • dan815
do u get why the vertex for y=a(x-b)^2+c is at (b,c) then?
anonymous
  • anonymous
dan815
  • dan815
okay then y=4x^2 + 24x + 32 rewrite this in that form
anonymous
  • anonymous
y=a(x-b)^2+c y=4(x-24)^2+32
dan815
  • dan815
thats not how it works
dan815
  • dan815
expand the brackets see if u get the right equation
anonymous
  • anonymous
y=x^2+6x+8
anonymous
  • anonymous
ohhhh
dan815
  • dan815
y=4(x-24)^2+32 =4*(x-24)(x-24)+32 = 4*(x^2-48x+24^2)+32
dan815
  • dan815
its much bigger than before now
dan815
  • dan815
so we have to transform to the form we want properly
anonymous
  • anonymous
okay so you did the square
dan815
  • dan815
okay ill show u how to do it this time, u can try to do it later
anonymous
  • anonymous
(a+b)^2=a^2+2ab+b^2
anonymous
  • anonymous
= 4*(x^2-48x+24^2)+32
dan815
  • 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
  • anonymous
thank you so much
anonymous
  • anonymous
uhhhhh finally i understand
dan815
  • dan815
y=4x^2 + 32x + 32 tell me the vertex for this equation and i will believe you
dan815
  • dan815
follow the steps i went through up there, its a very similar question
anonymous
  • anonymous
@dan815 whoa you went thru all at
dan815
  • 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
  • dan815
u can do these in ur head pretty much
anonymous
  • anonymous
yea ikr thats why i said that lol
anonymous
  • anonymous
y=4x^2 + 32x + 32 y=4(x^2+8x)+32 would it be that
anonymous
  • anonymous
so i can continue
dan815
  • dan815
thats a start now u have to complete the square in those brackets go on
anonymous
  • 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
  • anonymous
:w
dan815
  • dan815
no
dan815
  • dan815
the question u are asking urself here is how can i get a perfect square out of x^2+8x
dan815
  • 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
  • anonymous
so you have to multiply 8 by 2
dan815
  • dan815
u are not multiplying 8 by 2
dan815
  • dan815
u are dividing (8/2) then squaring (8/2)^2 = 16 just so happesn that its also the same as 8*2
anonymous
  • 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
  • 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
  • dan815
yes now u got it so what is the vertex
anonymous
  • anonymous
-4,-4??
dan815
  • dan815
no
dan815
  • dan815
what is the constant term
anonymous
  • anonymous
16
dan815
  • dan815
=4*(x+4)^2+32-16*4 simplfy this to a(x-b)^2+c
dan815
  • dan815
=4*(x+4)^2+32-16*4 =4*(x+4)^2+32-64 =??
anonymous
  • anonymous
-32
anonymous
  • anonymous
=4*(x+4)^2-32
dan815
  • dan815
oaky so what is the vertex?
anonymous
  • anonymous
i keep thinking its -4,-3
anonymous
  • anonymous
but i don't think it is that
dan815
  • dan815
what is the vertex for this equation =a(x-b)^2+c
anonymous
  • anonymous
b,c
dan815
  • dan815
okay so what is the vertex for this =4*(x+4)^2-32
anonymous
  • anonymous
-4,-32
dan815
  • dan815
yes
anonymous
  • anonymous
wow
dan815
  • dan815
:)
anonymous
  • anonymous
that was really hard but now i got it lol
anonymous
  • anonymous
only the inside changes sign, its the opposite
anonymous
  • anonymous
thank you dan815
anonymous
  • anonymous
you're a lifesaver!
dan815
  • dan815
no problem, try some completing the square excercises just to make sure though
dan815
  • dan815
or just do some expanding excercised, that should build your intuition up
anonymous
  • anonymous
i tried doing it with analytic geometry method lol
anonymous
  • anonymous
thank you
anonymous
  • anonymous
works too
dan815
  • dan815
dantheman
dan815
  • dan815
for the record here is a short cut
anonymous
  • anonymous
would it be the same for Writing the quadratic function in vertex form. y = x2 + 8x + 18
dan815
  • 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
  • anonymous
ohh wow that as much faster
dan815
  • dan815
NO do not use it
dan815
  • dan815
since u do not understand it
dan815
  • dan815
not until you understand completing the square method, then u can derive this
dan815
  • dan815
and then u can derive it one more time after u learn about derivatives
anonymous
  • anonymous
yea i told him that stuff earlier lol -b/2a
dan815
  • 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
  • anonymous
thank you
dan815
  • dan815
hint complete square for f(x)=ax^2+bx+c
anonymous
  • anonymous
:)
anonymous
  • anonymous
how can i rate a qualified helper
anonymous
  • anonymous
(5 starts)
anonymous
  • anonymous
lol
Jhannybean
  • Jhannybean
Similar to my method without simplifying the equation first

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