# For the function y = x^2 - 2x - 5, determine a) Its Vertex b) The x-intercepts in simplified form

Hi, Kristen,

First, we recognize this is a quadratic with a U-shaped graph. There is a formula that exists for the **x-coordinate for the vertex**:

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

Where a and b come from the equation y = ax^2 + bx + c. Here, a = 1, b = -2. So, we have:

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

Again, that is the x coordinate of the vertex. Whenever we have any x, to find the y, we can plug it into the original equation. So:

**y = (-1)^2 - 2(-1) - 5 = 1 + 2 - 5 = -2**

So, the vertex is **(-1,-2)**

**For all x-intercepts, y = 0**. So, for this, it would be exactly like solving the equation:

**0 = x^2 - 2x - 5**

This doesn't factor. So, we have to use the quadratic formula or complete the square. Doing complete the square, first, add 5 to each side. So:

**5 = x^2 - 2x**

Then, take half of b, -2/2 = -1. Then, square that result, (-1)^2 = 1. Add that result to both sides.

**5+1 = x^2 - 2x + 1**

The right side factors now:

**6 = (x-1)^2**

Square root each side, then add 1 to each side, we have:

**x = 1 +- sqrt(6)**

So, the x intercepts are **(1 + sqrt 6, 0) and (1-sqrt 6, 0)**.

Good luck, Kristen. I hope this helps.

Till Then,

Steve

If the equation of a parabola y = ax^2 + bx + c is written in the form y = a(x - h)^2 + k, the vertex of the parabola is the point (h, k).

For the parabola y = x^2 - 2x - 5, to determine the vertex write the equation in the form above:

y = x^2 - 2x - 5

y = x^2 - 2x + 1 - 6

y = (x - 1)^2 - 6

This gives the vertex as (1, -6)

At the points where the parabola intersects the x-axis the value of y is 0.

Solve the equation x^2 - 2x - 5 = 0

x^2 - 2x + 1 - 6 = 0

(x - 1)^2 = 6

`x - 1 = +- sqrt6`

`x = 1 +- sqrt6`

The x-intercepts of the parabola are `x +- sqrt6`