Your function is not defined at x=0 and x=4 (you have vertical asymptotes there). So, the denominator must be zero at x=0 and x=4. This is the case for

So, the function with that denominator is graphed.
The equation in the vertex form of g(x) is:

<h3>
How to get the equation of function g?</h3>
We know that g(x) is a translation of 2 units below and 3 units to the left of f(x).
So first, let's rewrite f(x) to its vertex form:

The vertex is at:

The y-value of the vertex is:

Then the vertex form of f(x) is:

If we move this vertex 2 units below, and 3 units to the left, then we have:

That is the equation for g(x) in vertex form.
If you want to learn more about quadratic equations:
brainly.com/question/1214333
#SPJ1
A=100°
as 180-140= 40 , 40 x 2 = 80 & 180-80 = 100° = a
Answer:

Step-by-step explanation:
If the population increases at a rate of 4% per annum, then:
In year 1:

Where
is the initial population and
is the population in year n
In year 2

It can also be written as:

Taking out common factor

Taking out common factor (1 + 0.04)

Taking out again common factor 
Simplifying

So

This is the equation that represents the population for year n
Then, in 4 years, the population will be:

Answer:
?????.... hii erica herrrr