Match each quadratic function with its respective graph. f (x) = x^2 + 6x - 5
f (x) = x^2 - 4x + 5
f (x) = -x^2 - 2x +3
f (x) = x^2 + 4x - 5
f (x) = -x^2 + 2x - 3
f (x) = x^2 - 6x + 5

The quadratic functions that fits with its respective graph are.

1). f (x) = x^2 - 4x + 5

2). f (x) = -x^2 - 2x +3

3). f (x) = x^2 - 6x + 5

Here is how it looks.

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: