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Answer: Choice A</h3>
- Domain: x > 4
- Range: y > 0
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Explanation:
We want to avoid having a negative number under the square root. Solving
leads to 
So it appears the domain could involve x = 4 itself; however, if we tried that x value, then we'd get a division by zero error.
So in reality, the domain is x > 4.
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The range of y = sqrt(x) is the set of positive real numbers. So y > 0 is the range for this equation. Shifting left and right does not affect the range, so the range of y = sqrt(x-4) is also y > 0.
We are dividing a positive number (3) over some positive number in the denominator. Overall, the expression
is positive because positive/positive = positive.
Therefore, the range of the given equation is y > 0
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The graph is shown below. We have a vertical asymptote at x = 4 and a horizontal asymptote at y = 0. The green curve is fenced in the upper right corner (northeast corner).
Treat this as you would the quadratic equation x^2 - 4x - 3 + 0. Solve this by completing the square:
x^2 - 4x + 4 - 4 - 7 = 0
(x^2 - 4x + 4) = 11
(x-2)^2 = 11, and so x-2 = plus or minus sqrt(11).
Graph this, using a dashed curve (not a solid curve). Then shade the coordinate plane ABOVE the graph.