Question

How do you find the VERTEX of a parabola `y=x^2-x-6`?

Answer 1

The vertex is `(1/2-25/4)` or `(1/2, -6 1/4)`.

Explanation:

`y=x^2-x-6` is a quadratic equation in the form of `ax^2+b^2+6`, where `a=1, b=-1, and c=-6`.

The vertex is the minimum or maximum point of the equation. The `x` value can be found using the formula `x=(-b)/(2a)`. The `y` value can be found by substituting the value for `x` into the equation and solving for `y`.

The value of `x`.

`x=(-b)/(2a)=(-(-1))/(2*1)=1/2`

`x=1/2`

The value of `y`.

`y=(1/2)^2-1(1/2)-6` =

`y=1/4-1/2-6` =

The common denominator for the right side is `4`. Multiply each term by a fraction so that it has a denominator of `4`.

`y=1/4-1/2(2/2)-6(4/4)` =

`y=1/4-2/4-24/4` =

`y=-25/4`

Simplify the improper fraction.

`y=-6 1/4`

graph{y=x^2-x-6 [-14.24, 14.23, -7.12, 7.12]}