Posted by **ladybug** on Monday, November 12, 2012 at 10:19pm.

I need to find the vertex and axis of symmetry of the folowing quadratic function. Can someone PLEASE help me!!

y= -x^2-6x

- college algebra--need help Please!!! -
**Reiny**, Monday, November 12, 2012 at 11:17pm
You can complete the square ....

y = -(x^2 + 6x **+9 - 9**

= -( (x+3)^2 - 9)

= -(x+3)^2 + 9

vertex is (-3, 9)

alternate method:

for y = ax^2 + bx + c

the x of the vertex is -b/2a

for for yours

the x of the vertex is -(-6)/-2 = -3

sub into the function ...

y = -9 + 18 = 9

so the vertex is (-3,9) as above

The simplest way is to use Calculus, but I am not sure if you are studying Calculus.

- college algebra--need help Please!!! -
**ladybug**, Tuesday, November 13, 2012 at 3:51am
Reiny,

It seems like Calculus, but it actually falls under college algebra. Because I can get some calculus tutoring and the answers fit. Thank you for the help, it is greatlt appreciated!!!

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