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anonymous
 one year ago
!!!
anonymous
 one year ago
!!!

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campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1here is a simple method in a quadratic \[y = ax^2 + bx + c\] the line of symmetry is found using \[x = \frac{b}{2a}\] so in your question a = 3 and b = 6 so find the line of symmetry now the vertex always lies on the line of symmetry, so substitute you answer into the equation to find the value of y.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1the alternative is to factor out 3 \[y = 3(x^2 + 2x + 1) \] now factor

Pawanyadav
 one year ago
Best ResponseYou've already chosen the best response.0Coordinates are (b/2a,d/4a) b=6 a=3 d=b^24ac=6^24×3×6=36 Now put values in coordinates of vertax

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I did dw:1434229051281:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then after I plugged in 1 for x in the equation 3(1^2)+6(1)+3=6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the coordinates would be (1,6) ?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well x= 1 is correct... so you substitution has an error \[f(1) = 3 \times (1)^2 + 6 \times (1) + 3 = 0\] so y = 0 and x = 1 then the vertex is (1, 0) if you factor \[Y = 3(x^2 + 2x + 1) ~~~becomes~~~~y=3(x + 1)^2\] so comparing it to \[y = a(x  h)^2 + k\] h = 1 and k = 0 which is what you go

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I see where I made the mistake, I was supposed to do it 3(1)^2. Thank you for your help, again!
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