How do you find the axis of symmetry and the vertex for #y = -x^2 -10x?#

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Answer 1 · 2017-03-24T07:21:52

Axis of symmetry: #x = -5#

The vertex is at #(-10, 25)#

There is a formula to find the axis of symmetry of a parabola.
The standard form of a parabola is #y = ax^2 +bx +c#

Axis of symmetry: #x = (-b)/(2a)#

We have #y = -x^2 -10x #

#a = -1 and b= -10#

#x = (-(-10))/(2(-1)) = 10/-2 = -5#

#x=-5# is the line of symmetry.

The vertex lies on the line of symmetry, so as soon as you have the #x#-value, you can find the #y#-value.

#y= -(-5)^2 -10(-5)#

#y= -25+50#

#y =25#

The vertex is at #(-10, 25)#