If the domain of the coordinate transformation f(x, y)=(y+2,-x-4) is (1,-4),(3,-2), (0,-1)

The Fundamental Graphing Principle for Functions says that for a point (a, b) to be on the graph,

f(a) = b. In particular, we know f(0) = 1, f(2) = 3, f(4) = 3 and f(5) = 5. Suppose we wanted to

graph the function defined by the formula g(x) = f(x) + 2. Let’s take a minute to remind ourselves

of what g is doing. We start with an input x to the function f and we obtain the output f(x).

The function g takes the output f(x) and adds 2 to it. In order to graph g, we need to graph the

points (x, g(x)). How are we to find the values for g(x) without a formula for f(x)? The answer is

that we don’t need a formula for f(x), we just need the values of f(x). The values of f(x) are the

y values on the graph of y = f(x). For example, using the points indicated on the graph of f, we

can make the following table.

x (x, f(x)) f(x) g(x) = f(x)

In general, if (a, b) is on the graph of y = f(x), then f(a) = b, so g(a) = f(a) + 2 = b + 2. Hence,

(a, b+ 2) is on the graph o

main role x (x, f(x)) f(x) g(x) = f(x) + 2 (x, g(x))

0 (0, 1) 1 3 (0, 3)

2 (2, 3) 3 5 (2, 5)

4 (4, 3) 3 5 (4, 5)

5 (5, 5) 5 7 (5, 7)

Step-by-step explanation:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: