Question

Consider the inverse function
f-1(x) = - VX-2
Which conclusions can be drawn about f(x) = x2 + 2? Select three options.
f(x) has a limited range.
f(x) has a restricted domain.
f(x) has an x-intercept of (2,0).
f(x) has a maximum at the point (0, 2).
f(x) has a y-intercept at the point (0, 2).

Answer 1

A. True. The range of f(x) is $y \ge 2$ meaning that y cannot be smaller than 2.

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B. True. Normally $f(x) = x^2+2$ is not restricted in the domain as any real number will work for x, but we apply a restriction on f(x) so the inverse $f^{-1}(x)$ is possible. This is so the graph of f(x) is one-to-one; otherwise, the graph of f(x) fails the horizontal line test.

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C. False. The graph of f(x) is entirely above the x axis, so there are no x intercepts. Alternatively, plug x = 2 into f(x) and you'll find that f(2) does not equal 0.

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D. False. The leading coefficient is positive, so the parabola opens upward. Consequently, this means that there is a lowest point (aka minimum) but there is no maximum point as the graph of f(x) goes upward forever.

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E. True. Plug in x = 0 and you'll get f(0) = 2. This is where the curve crosses the y axis.

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To summarize, the three answers are: A, B, and E

Answer 2

Answer: A,B,E

Step-by-step explanation: