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anonymous
 5 years ago
verify that the xcoordinate of the vertex of a parabola of the form y= ax^2 +bx +c is b/2a.
anonymous
 5 years ago
verify that the xcoordinate of the vertex of a parabola of the form y= ax^2 +bx +c is b/2a.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0set it equal to zero and solve for x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i have to solve it using 'completing the squares'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wat do u mean by that?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0f(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))^2b/a so the vertex is (b/[2a],b/a)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0oops i left something out

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0f(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])^2b/[2a] The vertex is (b/[2a],b/[2a])

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0lol oops that 4th step should be =a(x+b/[2a])^2b^2/[4a] the vertex is (b/[2a],b^2/[4a])

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0"verify that the xcoordinate 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
 5 years ago
Best ResponseYou've already chosen the best response.0we are learning 'cpmpleting the sqaures'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets step thru it then by "completeing the square"......cue thunder clap........what?!?...no thunder?..........oh well, moving along.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ax^2 + bx +c = 0 lets solve for any quadratic equation by going thru this setup..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0a 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok.... 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
 5 years ago
Best ResponseYou've already chosen the best response.0hmm..we divide the middle term by 2 ...then square it and stuff...yup

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0good good good.... :) but even on a more basic note.... if we add "0" to this equation do we change its value?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I think you do :) tell me: does (x^2 +6x +3) = (x^2 +6x +3 +0) ??

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0good :) 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: 99 to this equation, does the value change? remember: 99 = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cool... so we do this: (x^2 +6x +9) 9 +3 in order to "complete" the square right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so hw do we apply that to this q:

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0good... 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then 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
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0now... do you see the answer or can i do my magic and show you whats right before your eyes :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that 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
 5 years ago
Best ResponseYou've already chosen the best response.0put 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
 5 years ago
Best ResponseYou've already chosen the best response.0does that make sense for you?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0a parabola that is sagging beneath the x axis has its vertex halfway between its x intercepts..... its a giant "U".

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0b/2a is halfway between the intercept on the left and the intercept on the right...... that makes it the vertex.
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