Find the equation of the quadratic function with zeros 10 and 12 and vertex at (11, -2). y = 2x2 - 44x + 120 y = 2x2 - 44x - 240 y = 2x2 - 44x + 240 y = -2x2 - 44x + 240 im thinking its either the last one or second the the last one.

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Find the equation of the quadratic function with zeros 10 and 12 and vertex at (11, -2). y = 2x2 - 44x + 120 y = 2x2 - 44x - 240 y = 2x2 - 44x + 240 y = -2x2 - 44x + 240 im thinking its either the last one or second the the last one.

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vertex form is a(x - h)^2 + k where (h,k) are the coordinates the the vertex
and a is a number to be found
how do i find it?

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so here h = 11 and k = -2
y = a(x - 11)^2 - 2 plug in one of the zeroes say y = 0 and x = 10 and solve for a
0=a(10-11)^2-2 0=-1a a=0
your secon ine is wrong
(10-11)^2 = -1^2 = 1
okay.
0 = a - 2 a = 2
so now you need to expand y = 2(x - 11)^2 - 2
and that is done by..? sorry...i rly dont get this stuff..
2(x - 11)^2 - 2 = 2[x^2 -22x + 121] - 2 can you continues
multiply each term in the parentheses by 2
oh okay i see!
so now i expand it
yes
2(x-11) (x-11)-2
no you just multiply 2x^2 - 22x + 121 by 2 then take 2 away
what? im sorry im confused
- ive worked out (x - 11)(x - 11) already
(x - 11)(x - 110 = x^2 - 22x + 121
* 11) not 110
okay.. so i multiply the whole thing by 2?
yes -then take 2 away ( the -2 at the end)
4x^2-44x+242. now subract 2 from each number?
no - just from the 242 you can only subtract like terms
4x2-44+240
then take away a 2?
oh its 2*x2 = 2x^2 not 4x^2
okay i was gonna say. lol everything else matches but the 4x lol
2x^2 - 44x + 242 - 2 = 2x^2 - 44x + 240
got it! so i was close, i had a feeling it be that one or the last one.

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