Question

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

Answer 1

how do i put it in vertex form

Answer 2

you need to complete the square.

first take out 4 as a common factor:

4x^2 + 24x + 32 = 4(x^2 + 6x + 8)

6/2 = 3 so:

= 4( (x+3)^2 - 3^2 + 8 )

= 4( (x+3)^2 - 9 + 8)

= 4( (x+3)^2 - 1) = 4(x+3)^2 - 4


vertex will be the minimum point, so:

now, 4(x+3)^2 - 4 is at a minimum when x = -3 because it makes the stuff being squared equal to 0
so we know the x coord of vertex is -3

the y coord is the constant on the end, -4

Answer 3

does that make sense?

Answer 4

sorry ill explain a bit more :) anything that confuses you dont be worried about saying. im not very good at explaining so it'll probably be my fault

Answer 5

4x^2 + 24x + 32

the bit about the 4 being a common factor isnt that important, but it makes things a bit easier.

i realized it was a common factor because 4x^2 is 4*(x^2) ,
24x is 4*(6x)
and 32 is 4*(8)

can you see that they are all multiples of 4? don't worry if you don't or if you take a lot of time to, it all comes with experience.

Answer 6

oh i seee.... okay and then next you put 4(x+3)^2-3^2+8 can you explain what you did in the next 3 steps

Answer 7

sure :)

Answer 8

4x^2 + 24x + 32 = 4(x^2 + 6x + 8)

lets just forget about the 4 for a minute, and we'll remember him later
we want to "complete the square" for:

x^2 + 6x + 8

the basic idea is this
look at this:
(x+3)^2 = (x+3)(x+3) = x^2 + 3x + 3x + 9

= x^2 + 6x + 9

this is almost what we have isn't it?

Answer 9

Are you real teachers?

Answer 10

if

(x+3)^2 = x^2 + 6x + 9

then (x + 3)^2 - 1 = (x^2 + 6x + 9) -1 = x^2 + 6x + 8

so (x + 3)^2 - 1 = x^2 + 6x + 8

the reason (x+3) works is because 3 is half of 6

Answer 11

@headge im not a teacher, im a student

Answer 12

WOOOOOW

Answer 13

Eigenschmeigen?

Answer 14

@eigenschmeigen ?

Answer 15

?