How do you find the axis of symmetry, vertex and x intercepts for #y=x^2-2x-1#?

1 Answer
Oct 24, 2017

Axis of symmetry: #x=1#

Vertex: #(1,-2)#

X-intercepts: #(1+sqrt2,0), ##(1-sqrt2,0)#

Explanation:

Given:

#y=x^2-2x-1# is a quadratic equation in standard form:

#ax^2+bx+c#,

where:

#a=1#, #b=-2#, and #c=-1#

Axis of Symmetry: the line that divides the parabola into two equal halves

To find the axis of symmetry, #(x,0)#, use the formula:

#x=(-b)/(2a)#

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

#x=2/2=1#

Axis of symmetry: #(1,0)#

Vertex: the minimum or maximum value of a parabola

Substitute #1# for #x# in the equation and solve for #y#.

#y=1^2-2*1-1#

Simplify.

#y=1-2-1#

#y=-2#

Vertex: #(1,-2)#, which is the minimum point of the parabola and the parabola opens upward.

X-intercepts: values for #x# when #y=0#

Substitute #0# for #y#.

#0=x^2-2x-1#

Use the quadratic formula to solve for the x-intercepts.

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Plug in the known values.

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

Simplify.

#x=(2+-sqrt8)/2#

Simplify #sqrt8#.

#x=(2+-((2xx2)xx2))/2#

#x=(2+-2sqrt2)/2#

Reduce.

#x=[(2+-2sqrt2)/2]/2#

#x=1+-sqrt2#

X-intercepts: #(1+sqrt2,0), ##(1-sqrt2,0)#

approximate values: #(2.414,0),##(-0.4142,0)#

Summary

Axis of symmetry: #x=1#

Vertex: #(1,-2)#

X-intercepts: #(1+sqrt2,0), ##(1-sqrt2,0)#

graph{x^2-2x-1 [-11.25, 11.25, -5.625, 5.625]}