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

1 Answer
EZ as pi
Sep 5, 2016

Axis of sym at #x= -1#, vertex at (-1,1)

#x-#intercepts at -2 and 0

#y#-intercept at (0,0)

Explanation:

Let's find the axis of symmetry first, it is an easy way to find the vertex.
#color(red)("axis of symmetry:" x= (-b)/(2a)) = (-(-2))/(2(-1)) = -1#
#x=-1#

The vertex lies on the axis of symmetry:

If #x=-1, " find " y = -(-1)^2 -2(-1) = -1+2 = 1#

#color(blue)("The vertex is at " (-1,1))#

Intercepts:
For y-int: #x=0 rarr y=0#

For x-intercepts : # y = 0#

#0 = -x^2 -2x" "larr # factorise

#0 = -x(x+2) rArr x=0, or x=-2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the completing the square method to find the vertex:

#y = -x^2 -2x#
#y = -1(x^2 +2x)#
#y=-1[color(magenta)(x^2 +2x +1)-1]#
#y =-1[color(magenta)((x+1)^2) -1]#
#y = -(xcolor(blue)(+1))^2 color(blue)(+1)#

#color(blue)("The vertex is at " (-1,1))#

Confirm these points on the graph below.

graph{-x^2-2x [-3.5, 1.5, -1.4, 1.1]}

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