Question

verify that the x-coordinate of the vertex of a parabola of the form y= ax^2 +bx +c is -b/2a.

Answer 1

differentiate

Answer 2

set it equal to zero and solve for x

Answer 3

but i have to solve it using 'completing the squares'

Answer 4

so no calc?

Answer 5

wat do u mean by that?

Answer 6

f(x)=ax^2+bx+c=a(x^2+bx/a)+c=
a(x^2+bx/a+(b/[2a])^2)-a(b/a)^2
=a(x+b/(2a))^2-b/a
so the vertex is (b/[2a],-b/a)

Answer 7

oops i left something out

Answer 8

f(x)=ax^2+bx+c
=a(x^2+bx/a)+c
=a(x^2+bx/a+(b/[2a])^2)-a(b/[2a])^2
=a(x+b/[2a])^2-b/[2a]
The vertex is (-b/[2a],-b/[2a])

Answer 9

lol oops that 4th step should be
=a(x+b/[2a])^2-b^2/[4a]
the vertex is (-b/[2a],-b^2/[4a])

Answer 10

this one?

Answer 11

ya i dont get it

Answer 12

"verify that the x-coordinate of the vertex of a parabola of the form y= ax^2 +bx +c is -b/2a."

Ok. this is really a part of the quadratic formula. Do you know the quad formula?

Answer 13

nop

Answer 14

we are learning 'cpmpleting the sqaures'

Answer 15

lets step thru it then by "completeing the square"......cue thunder clap........what?!?...no thunder?..........oh well, moving along.

Answer 16

ax^2 + bx +c = 0

lets solve for any quadratic equation by going thru this setup..

Answer 17

k

Answer 18

a little history tho....

a "complete" square looks like this:

(x+3)^2 is a complete square...... when we expand it we get something that looks like this:

x^2 +6x +9.. right?

Answer 19

ya

Answer 20

ok.... then if we can get an equation to be a "complete" square, then solving it is easier...

take the equation:

x^2 +6x +3. this looks similar, but it is not a "complete" square is it?

Answer 21

hmm..we divide the middle term by 2 ...then square it and stuff...yup

Answer 22

good good good.... :)

but even on a more basic note.... if we add "0" to this equation do we change its value?

Answer 23

i dont know

Answer 24

I think you do :)

tell me:

does (x^2 +6x +3) = (x^2 +6x +3 +0) ??

Answer 25

ya

Answer 26

good :)

then what we really need is a convenient form of "0" to apply to this equation.

we know we want a +9 in there somewhere right? so what happens if we add:

9-9 to this equation, does the value change?

remember: 9-9 = 0

Answer 27

s

Answer 28

s?

Answer 29

i get wat u r saying

Answer 30

cool...

so we do this:

(x^2 +6x +9) -9 +3 in order to "complete" the square right?

Answer 31

ya

Answer 32

so hw do we apply that to this q:

Answer 33

good...

and you already pointed out the (b/2)^2 is what we will need given any equation of this form correct?

(6/2)^2 = 3^2 = 9 right

Answer 34

yup

Answer 35

then lets begin to see what this form of equation is hiding from us :)

ax^2 +bx + c = 0 ; subtract c from both sides...

ax^2 + bx = -c ; divide at all by "a"

x^2 +(b/a)x = (-c/a) ; complete the square

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

Answer 36

i hope your still with me on this :)

(x +(b/2a))^2 = (-c/a) + (b^2/4a^2) ; lets add the right side together with the rules for add fractions....

(x +(b/2a))^2 = (b^2 -4ac)/2a ; now we squareroot both sides.

x+ (b/2a) = +- sqrt(b^2 -4ac)/2a ; and to solve for "x" we subtract that (b/2a) and get...

x = -b/2a +- sqrt(b^2 -4ac)/2a we good up to this point? cause this is where the question gets answered ;)

Answer 37

now... do you see the answer or can i do my magic and show you whats right before your eyes :)

Answer 38

ya

Answer 39

that right side part...that +- sqrt(...)/2a is just a number..... lets go ahead and make it equal to, i dunno, 6

x = -b/2a (+or-) "6"

what does this tell ya?

<.......-6.........-b/2a..........+6...........>

-b/2a sits right smack between this 6's right? its can be nothing else BUT the vertex

Answer 40

put on your -b/2a shirt stand up and look to your left for 6 feet and look to the right for 6 feet.... are you in the middle?

Answer 41

does that make sense for you?

Answer 42

a parabola that is sagging beneath the x axis has its vertex halfway between its x intercepts..... its a giant "U".

Answer 43

-b/2a is halfway between the intercept on the left and the intercept on the right...... that makes it the vertex.