## anonymous one year ago !!!

1. 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.

2. campbell_st

the alternative is to factor out 3 $y = 3(x^2 + 2x + 1)$ now factor

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

4. anonymous

I did |dw:1434229051281:dw|

5. anonymous

Then after I plugged in -1 for x in the equation 3(-1^2)+6(1)+3=6

6. anonymous

So the coordinates would be (-1,6) ?

7. 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

8. anonymous

Oh I see where I made the mistake, I was supposed to do it 3(1)^2. Thank you for your help, again!