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

Mathematics
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differentiate
set it equal to zero and solve for x
but i have to solve it using 'completing the squares'

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so no calc?
wat do u mean by that?
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)
oops i left something out
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])
lol oops that 4th step should be =a(x+b/[2a])^2-b^2/[4a] the vertex is (-b/[2a],-b^2/[4a])
this one?
ya i dont get it
"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?
nop
we are learning 'cpmpleting the sqaures'
lets step thru it then by "completeing the square"......cue thunder clap........what?!?...no thunder?..........oh well, moving along.
ax^2 + bx +c = 0 lets solve for any quadratic equation by going thru this setup..
k
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?
ya
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?
hmm..we divide the middle term by 2 ...then square it and stuff...yup
good good good.... :) but even on a more basic note.... if we add "0" to this equation do we change its value?
i dont know
I think you do :) tell me: does (x^2 +6x +3) = (x^2 +6x +3 +0) ??
ya
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
s
s?
i get wat u r saying
cool... so we do this: (x^2 +6x +9) -9 +3 in order to "complete" the square right?
ya
so hw do we apply that to this q:
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
yup
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
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 ;)
now... do you see the answer or can i do my magic and show you whats right before your eyes :)
ya
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
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?
does that make sense for you?
a parabola that is sagging beneath the x axis has its vertex halfway between its x intercepts..... its a giant "U".
-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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