# For the function y = x^2 - 2x - 5 determine: a) it's vertex. [use (,) with coordinates] b) the x intercepts in simplified form.``

*print*Print*list*Cite

### 2 Answers

The given function is `y = x^2 - 2x - 5.`

This is the equation of a parabola in its standard form.

To find its vertex transform the given function into the equation of the parabola in its vertex form i.e `y=a(x-h)^2+k` where `(h,k)` is its vertex.

So, `y = x^2 - 2x - 5`

`rArr y = x^2 - 2x +1-1- 5`

`rArr y=(x-1)^2-6`

Here, `h=1` , `k=-6`

**a)** **Hence, the vertex of the given function is (1,-6)**.

To find the x intercepts set `y=0` and solve. So,

`x^2 - 2x - 5=0`

Applying the quadratic formula we get:

`x=(2+-sqrt(2^2-4*1*-5))/2`

`rArr x=(2+-sqrt(24))/2`

`rArr x=(2+-2sqrt6)/2`

`rArr x=1+-sqrt6`

**b) Therefore, the x intercepts are** `1+sqrt6` and `1-sqrt6` .

The graph:

**Sources:**