How do you find the axis of symmetry, vertex and x intercepts for #y=x^2+6x+5#?

1 Answer
Apr 12, 2017

Axis of symmetry: #x=-3#
Vertex: #(-3,-4)#
X-intercepts: #x=-5# and #x=-1#

Explanation:

Axis of symmetry:-
Can be found by applying #x=-b/(2a)#
In the equation #y=color(red)1x^2+color(blue)6x+color(green)5, a=1, b=6, c=5#
So, #x=-6/(2(1))=color(purple)(-3)#

Vertex:-
we substitute the #x# value #(color(purple)(-3))# we found in the axis of symmetry in the function, and we will get
#(-3)^2+6(-3)+5=9-18+5=color(magenta)(-4)#,
The vertex is the point #(color(purple)(-3), color(magenta)(-4))#

X-intercepts:-
X-intercept is when #y=0#, which means #x^2+6x+5# should be equal to zero.
#x^2+color(orange)6x+color(grey)5=0#
Now we can factor it by this way since the coefficient of #x^2# is #1#, what two numbers if multiplied will give you #color(grey)5# and if added will give you #color(orange)6#? They are #color(brown)5# and #color(darkblue)1#, so,
#(x+color(brown)5)(x+color(darkblue)1)=0#
#x+5=0# and #x+1=0#
#x=-5# and #x=-1#