How do you find the vertex and axis of symmetry, and then graph the parabola given by: #y= x^2 + 6x + 5#?

#y=x^2+6x+5# is a quadratic equation which has the form #y=ax^2+bx+c#, in which #a=1, b=6, and c=5#.

Axis of Symmetry

The axis of symmetry is determined by the formula #x=(-b)/(2a)#.

#x=(-6)/(2*1)=-6/2=-3#

The axis of symmetry is #x=-3#

Vertex

The vertex is the point #(x,y)# that is the maximum or minimum of a parabola. Since #a# is positive, the parabola for this graph will open upward and the vertex will be the minimum point.

The #x# value of the vertex is the same as the axis of symmetry. #x=-3#.

To find the #y# value of the vertex, substitute #-3# for #x# in the equation #y=x^2+6x+5#.

#y=(-3)^2+6(-3)+5=#

#y=9-18+5=-4#

The vertex is #(-3,-4)#.

Determine several points on both sides of the axis of symmetry by substituting values for #x# into the equation and solving for #y#.

#y=x^2+6x+5#

#x=-6,# #y=5#
#x=-5,# #y=0#
#x=-4,# #y=-3#
#x=-3,# #y=-4# (vertex)
#x=-2,# #y=-3#
#x=-1,# #y=0#
#x=0,# #y=5#

Plot the points and sketch an upward opening parabola with the curve at the vertex. Do not connect the dots.

graph{y=x^2+6x+5 [-16.02, 16, -8.01, 8.01]}