How do you find the axis of symmetry, vertex and x intercepts for #y=x^2+x-6#?

2 Answers
Mar 24, 2017

the vertex will be #(-1/2,-25/4)#
the axis of symmetry is simply #X=-1/2#

#x" "#intercepts #x=-3,2#

Explanation:

The first thing to do is to complete the square

#y=x^2+x-6#

add half the coefficient of #x# and square it, and subtract to balance

#y=color(red)(x^2+x+(1/2)^2)-6-(1/2)^2#

the terms in red give a perfect square

#y=color(red)((x+1/2)^2)-25/4#

the vertex will be #(-1/2,-25/4)#

basically this is the translation of #y=x^2# to this graph.

graph{x^2+x-6 [-10, 10, -5, 5]}

the axis of symmetry is simply #X=-1/2#

the #x# intercepts we solve

#(x+1/2)^2-25/4=0#

#=>(x+1/2)^2=25/4#

#x+1/2=+-sqrt(25/4)#

#x+1/2=+-sqrt(25/4)#

#x+1/2=+-5/2#

#:.x=+-5/2-1/2#

#x=-3,2#

Mar 24, 2017

Axis of symmetry: #x = -1/2#
Vertex is at #(-1/2, -6 1/4)#

#x#-intercepts are at #-3 and 2#

Explanation:

The quadratic is in the form #y = ax^2 +bx +c#

#y = x^2 +x -6" "a = 1, b= 1 and c = -6#

The axis of symmetry is found from: #x = (-b)/(2a)#

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

The vertex lies on the axis of symmetry.
You have the #x#-value, substitute to find the #y#-value:

#y = (-1/2)^2 -1/2-6" "= -6 1/4#

Vertex is at #" "(-1/2, -6 1/4)#

To find the #x#-intercepts, set #y=0# and solve

#x^2 +x-6 =0" "larr# factorise

#(x+3)(x-2)=0#

#x = -3 and x = +2#

(The average of the x-intercepts confirms the axis of symmetry)

#x = (-3+2)/2 = -1/2#