anonymous
  • anonymous
!!!
Mathematics
jamiebookeater
  • jamiebookeater
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campbell_st
  • campbell_st
here 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
  • campbell_st
the alternative is to factor out 3 \[y = 3(x^2 + 2x + 1) \] now factor
Pawanyadav
  • Pawanyadav
Coordinates are (-b/2a,-d/4a) b=6 a=3 d=b^2-4ac=6^2-4×3×6=-36 Now put values in coordinates of vertax

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anonymous
  • anonymous
I did |dw:1434229051281:dw|
anonymous
  • anonymous
Then after I plugged in -1 for x in the equation 3(-1^2)+6(1)+3=6
anonymous
  • anonymous
So the coordinates would be (-1,6) ?
campbell_st
  • campbell_st
well 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
  • anonymous
Oh 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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