Question

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

Answer 1

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]}