C (see attached)
<h3>Step-by-step explanation:</h3>The linear portion of the graph is defined for x ≥ 1, with the point x=1 included. Only selections A and C do that.
The quadratic portion of the graph is defined for x < 1. Only selection C does that. (Selection A is doubly-defined for x > 1, so is not a function. It is undefined for x < 1.)
See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0
<h3>How to graph the hyperbola?</h3>The equation of the hyperbola is given as:
12x^2 - 3y^2 - 108 = 0
Start by calculating the transverse axis
So, we have:
<u>Transverse axis</u>
The vertices of the given hyperbola are (0, 0)
This means that
(h, k) = 0
Where
a = 3 and b = 6
The transverse axis is calculated as:
y = ±b/a(x - h) + k
So, we have:
y = ±6/3(x - 0) + 0
Evaluate the difference and sum
y = ±6/3x
Evaluate the quotient
y = ±2x
This means that the transverse axes are y = 2x and y =-2x
<u>The vertices</u>
In the above section, we have:
The vertices of the given hyperbola are (0, 0)
This means that
(h, k) = 0
<u>The co-vertices</u>
In the above section, we have:
The vertices of the given hyperbola are (0, 0)
This means that
(h, k) = 0
And
a = 3 and b = 6
The co-vertices are
(h - a, k) and (h + a, k)
So, we have:
(0 - 3, 0) and (0 + 3, 0)
Evaluate
(-3,0) and (3, 0)
See attachment for the graph of the hyperbola
Read more about hyperbola at:
#SPJ1
