How do you find the axis of symmetry and vertex point of the function: #y = -x^2 - x + 9#?

2 Answers
Jonny · MathFact-orials.blogspot.com
Jun 10, 2018

Vertex: #(-1/2,9 1/4)#
Line of symmetry: #x=-1/2#

Explanation:

use #-b/(2a)# to find the vertex:

#-(-1)/(2xx(-1)) = -1/2#

plug #-1/2# in for #x#

#= 9 1/4#

vertex: #(-1/2,9 1/4)#

on parabolic functions the axis of symmetry is the vertical line where the vertex is, so the equation is:

#x=-1/2#

MathFact-orials.blogspot.com
Jun 10, 2018

An alternate way, using the vertex form:

Explanation:

We can change the standard form of the equation to the vertex form, which has the general form:

#y=a(x-h)^2+k#

We get there by completing the square (we do that by taking the constant of the #x# term, halving it, squaring the half, then adding and subtracting it into the equation:

#y=-x^2-x+9#

#y=-(x^2+x)+9#

#y=-(x^2+x+1/4-1/4)+9#

#y=-((x+1/2)^2-1/4)+9#

#y=-(x+1/2)^2+1/4+9#

#y=-(x+1/2)^2+9 1/4#

The vertex is given by #(h,k)#, which in this case is #(-1/2, 9 1/4)#.

The axis of symmetry runs through the vertex vertically and so takes the form of #x= "the " x " value of the vertex"#, which in this case is

#x=-1/2#

This can all be seen in the graph:

graph{-x^2-x+9[-3,3,8,10]}

note that I've adjusted the graph unequally vertically and horizontally to better show the vertex

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