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

- anonymous

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

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- anonymous

differentiate

- anonymous

set it equal to zero and solve for x

- anonymous

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

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## More answers

- anonymous

so no calc?

- anonymous

wat do u mean by that?

- myininaya

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)

- myininaya

oops i left something out

- myininaya

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])

- myininaya

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

- amistre64

this one?

- anonymous

ya i dont get it

- amistre64

"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?

- anonymous

nop

- anonymous

we are learning 'cpmpleting the sqaures'

- amistre64

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

- amistre64

ax^2 + bx +c = 0
lets solve for any quadratic equation by going thru this setup..

- anonymous

k

- amistre64

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?

- anonymous

ya

- amistre64

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?

- anonymous

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

- amistre64

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

- anonymous

i dont know

- amistre64

I think you do :)
tell me:
does (x^2 +6x +3) = (x^2 +6x +3 +0) ??

- anonymous

ya

- amistre64

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

- anonymous

s

- amistre64

s?

- anonymous

i get wat u r saying

- amistre64

cool...
so we do this:
(x^2 +6x +9) -9 +3 in order to "complete" the square right?

- anonymous

ya

- anonymous

so hw do we apply that to this q:

- amistre64

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

- anonymous

yup

- amistre64

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

- amistre64

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 ;)

- amistre64

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

- anonymous

ya

- amistre64

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

- amistre64

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?

- amistre64

does that make sense for you?

- amistre64

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

- amistre64

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

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