Question

Find the vertex of the graph of the function.

f(x) = 4x^2 + 24x + 32

Answer 1

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: \(y=a(x-h)^2+k\)

Answer 2

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

Answer 3

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

Answer 4

a=4
b=24
c=32

Answer 5

y=a(x−h)2+k = = y=4(x−h)2+k

Answer 6

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

Answer 7

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

Answer 8

ohh and thank you so much for helping me @Jhannybean

Answer 9

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

Answer 10

@DanTheMan99

Answer 11

thank you

Answer 12

sorry mi wifi was acting up @Jhannybean

Answer 13

y=x2+6x+8

Answer 14

y=x2+6x+8 y=ax2+bx+c
? y=a(x−h)2+k

Answer 15

but how will i get the other values?

Answer 16

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?

Answer 17

not yet

Answer 18

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

Answer 19

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?

Answer 20

yes

Answer 21

do i get x and y intercepts?

Answer 22

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!

Answer 23

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.

Answer 24

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

Answer 25

so c = 9

Answer 26

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

Answer 27

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

Answer 28

y=x2+6x+9=−8+9
y=x2+6x+9=+1

Answer 29

ohh ok

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

Do you see how it works?

Answer 35

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

Answer 36

y=a(x−h)2+k⟶y=(x+3)2−1

Answer 37

⟶y=(x+3)^2−1

Answer 38

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

Answer 39

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

Answer 40

y=-1

Answer 41

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

Answer 42

Im not understanding that.

Answer 43

1 = (x−(−3))^2

Answer 44

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

Answer 45

To help you identify them a little better.

the part in red is your x-intercept, part in blue is your y intercept

Answer 46

thats much better, sorry i was confused

Answer 47

\[y=a(x−h)^{2}+k⟶y=(x−(−3))^{2}−1\]

Answer 48

wait so what do i do now? the intercepts?

Answer 49

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

Answer 50

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

Answer 51

Awesome! Glad I could help :)

Answer 52

@Jhannybean sorry to bother you again

Answer 53

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

Answer 54

im so confused :(

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

y=(x−(−3))^2−1

-\[\frac{ -3 }{ 1 }\]

Answer 59

-(-3/2)

Answer 60

right?

Answer 61

3

Answer 62

a and b is coefficient

Answer 63

all the numberse

Answer 64

24/8

Answer 65

add a negative sign

Answer 66

-3

Answer 67

yea

Answer 68

@saseal

Answer 69

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

Answer 70

lowest point is obviously the y-axis

Answer 71

im so lost

Answer 72

dantheman

Answer 73

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

Answer 74

GET A LIFE DAN SRS