Question

How do you find the vertex and the intercepts for `G(x) = x ^ 2 - 6`?

Answer 1

`x` intercepts at: `x=+-sqrt6`
`y` intercepts at the point `(0,-6)`.
Vertex at the point `(0,-6)`

Explanation:

Let's first have a look at the `x` intercepts of the function:

Recall that the difference between two squares can be expressed as: `n^2-m^2=(n-m)(n+m)`

If we consider 6 as:

`sqrt(6)^2`
We can now convert the function to a difference of squares:
`G(x)=x^2-6`
`G(x)=(x-sqrt6)(x+sqrt6)`

Using the Null Factor Law we can now solve for `x`, hence determine the `x` intercepts of the function:

`0=(x-sqrt6)(x+sqrt6)`

Therefore `x` intercepts are present at:
`x=+-sqrt6`

Let's now have a look at the `y` intercept of the function:

Consider that for any function, `y` intercepts where `x=0`. Therefore if we substitute `x=0` into the function, we can calculate the `y` intercept of the function.

`G(x)=x^2-6`
`G(x)=0^2-6`
`G(x)=-6`

Therefore, the `y` intercept of the function is present at the point `(0,-6)`.

Finally, let's have a look at the vertex of the function:

Consider the general form of a parabolic function:
`y=ax^2+bx+c`

If we compare the equation that you have presented:

`G(x)=x^2-6`

We can determine that:

The `x^2` coefficient is 1; this implies that `a=1`

The `x` coefficient is `0`; this implies that #b = 0#
The constant term is `-6`; this implies that
#c = -6#
Therefore, we can use the formula:
`Tp_x=−b/2a`

to determine the `x` value of the turning point.

Substituting the appropriate values into the formula we get:

`Tp_x=−0/(2*1)`
`Tp_x=−0/(2)`
`Tp_x=0`

Substituting this value into the given function, we get:

`G(x)=0^2-6`
`G(x)=0^2-6`
`G(x)=-6`

Therefore, the vertex of the function is present at the point `(0,-6)`.