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

### 2 Answers | Add Yours

Hi, Kristen,
To solve this, for the vertex, there is a formula we can use:
x = -b/(2a)
This gives the x coordinate of the vertex. a and b come from the general formula y = ax^2 + bx + c. Comparing your function to that formula, a = 1 and b = -2 (c = -5). So, plugging in:
x = -(-2)/(2*1) = 1
So, x = 1. To find the y, we can always plug the value of x back into the original equation. So, plugging in 1 for x:
y = 1^2 - 2*1 - 5 = 1 - 2 - 5 = -6
Therefore, the vertex is at (1,-6)
For the x intercepts, all x intercepts have y = 0. Therefore, we would actually be solving:
0 = x^2 - 2x - 5
It doesn't factor, so we have to solve using the quadratic formula or completing the square. To complete the square, we first make sure if a = 1. It is, so then, we move c to the other side. Here, we add 5:
5 = x^2 - 2x
Then, we take half of b, -2/2 = -1, then square that result, (-1)^2 = 1. We then add that result to each side:
5+1 = x^2 - 2x + 1
The right side factors
6 = (x-1)^2
Taking the square root of each side then adding 1 to each side:
1 +- sqrt 6 = x, or
x = 1 +- sqrt 6 = approx 3.45 and -1.45
So, the x intercepts would be (3.45,0) and (-1.45,0)
Good luck, Kristen. I hope this helps.
Till Then,
Steve

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:**