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

1 Answer
Apr 11, 2016

#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)#.