Need help finding the answer to this equation.

Question

I can't figure out this equation out. There are four choices but I don't know which is the right one. Please help.

f is a quadratic function whose graph is a parabola opening upward and has a vertex on the x-axis. The graph of the new function g defined by g(x) = 2 - f(x - 5) has a range defined by the interval

A. [ -5 , + infinity)

B. [ 2 , + infinity)

C. ( - infinity , 2]

D. ( - infinity , 0]

Answer 1

f is a quadratic function whose graph is a parabola opening upward and has a vertex on the x-axis. The graph of the new function g defined by g(x) = 2 - f(x - 5) has a range defined by the interval

A. [ -5 , + infinity)

B. [ 2 , + infinity)

C. ( - infinity , 2]

D. ( - infinity , 0]

Let's talk through this a one step at a time.

*Since f(x) is concave-up with its vertex on the x-axis, we know f(x) ≥ 0.

*We also know that when we shift a function's domain by a positive number, we shift the function left and when we shift a function's domain by a negative number, we shift the function right. So f(x-5) is f(x) shifted to the right by 5.

*At this point, f(x-5) has its vertex at (5,0).

*When we negate f(x-5), the parabola becomes concave down yet the vertex remains at (5,0). Now we're at -f(x-5). At this point we have -f(x-5)≤0 with a range (-∞,0]

*If we add 2 to create g(x)=2-f(x-5), then we have a concave down parabola with its vertex shifted up by 2, at (5,2). So, g(x) is concave down with its vertex at (5,2). Hence, its range is (-∞,2] which is choice C.

Answer 2

The function g(x) will have a range of (-∞, 2]. The function g(x) represents a transformation of the function f(x). The function g can be rewritten as -f(x - 5) + 2. The function f(x) has a range of [0, ∞) since its vertex is on the x-axis. The value of y at this point is 0.

The function g(x) states that f(x) has been shifted vertically upward two units, reflected over the x-axis (because it is being multiplied times -1), and shifted horizontally five units to the right. Note that the horizontal shift has no bearing on the range of the function.