Answer:
f(x) has a limited range
f(x) has a maximum at the point (0, 2)
f(x) has a y-intercept at the point (0, 2).
Step-by-step explanation:
Given the function;
f(x) = x^2+2
The domain is the value of the input variables for which the function will exist. According to the expression given, the function exists on all real values of x. The same goes with range which deals with the output values. It also exists on all real values from 2 and above.
Hence f(x) have a limited range (since values less than 2 are not included compare to domain that exists on all real values) and does not have a restricted domain.
For the x intercept, x intercept occur at y = 0
substitute y = 0 into the function and get y
if y = f(x)
y = x^2+2
0 = x^2 + 2
x^2 = -2
x = 2i
Hence f(x) does not have an x-intercept of (2, 0)
For the y intercept, y intercept occur at x = 0
substitute x = 0 into the function and get y
if y = f(x)
y = x^2+2
y = 0^2 + 2
y = 2
Hence f(x) has a y-intercept at point (0, 2)
f(x) is at maximum if d(fx))/dx = 0
d(fx))/dx = 2x
since d(fx))/dx = 0
0 = 2x
x = 0
substitute x = 0 into the function
f(x) = x^2 + 2
y = 0^2+2
y = 2
Hence f(x) has a maximum at the point (0, 2)
will have the form 
, where A and B are equivalent and factors that are multiplied in.
you will need to find a fraction which is a product of this: 