Which graph represents a proportional relationship?
1 answer:
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Answer:
range of f(x) = [-4, -2) ∪ [2, 8)
a+b+c+d = -4
Step-by-step explanation:
The graph is attached. The range is the vertical extent of the function. It is defined at f(0) = -4 and f(2) = 2.
The limits f(2-) and f(4-) are -2 and 8, respectively, so the graph has open circles there. These are the ends of the two half-open intervals that make up the range of the function.
The portion of the graph in the domain [4, 7) is included in the range [2, 8), so no special treatment is needed for that piece of the function.
Answer:
- x = ±√3, and they are actual solutions
- x = 3, but it is an extraneous solution
Step-by-step explanation:
The method often recommended for solving an equation of this sort is to multiply by the product of the denominators, then solve the resulting polynomial equation. When you do that, you get ...
... x^2(6x -18) = (2x -6)(9)
... 6x^2(x -3) -18(x -3) = 0
...6(x -3)(x^2 -3) = 0
... x = 3, x = ±√3
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Alternatively, you can subtract the right side of the equation and collect terms to get ...
... x^2/(2(x -3)) - 9/(6(x -3)) = 0
... (1/2)(x^2 -3)/(x -3) = 0
Here, the solution will be values of x that make the numerator zero:
... x = ±√3
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So, the actual solutions are x = ±3, and x = 3 is an extraneous solution. The value x=3 is actually excluded from the domain of the original equation, because the equation is undefined at that point.
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<em>Comment on the graph</em>
For the graph, we have rewritten the equation so it is of the form f(x)=0. The graphing program is able to highlight zero crossings, so this is a convenient form. When the equation is multiplied as described above, the resulting cubic has an extra zero-crossing at x=3 (blue curve). This is the extraneous solution.
