Q 1. Let the function be denoted by f(x). It may be observed from the graph that:
The line x = 4 appears to be a vertical asymptote so that (x-4) must be a factor of the denominator.
The graph shows a x-intercept at x = 3. Hence (x-3) must be a factor of the numerator. Also, y = 3/4 is a y-intercept so that (4x-3) must be a factor of the denominator.
In view of the above, the function of least degree satisying the given conditions is f(x) = a(x-3)/(4x-3)(x-4) where a is an arbitrary real number.
Note: If the line x = 4 is not a vertical asymptote, then f(x) = a(x-3)/(4x-3). The graph is not large enough for us to be sure about this.
Q. 2. Let the function be denoted by g(x).
The line x = 0 is a vertical asymptote so that x must be a factor of the denominator.
The x -intercept is -1 so that (x+1) must be a factor of the numerator.
Let g(x) = a(x+1)/x where a is an arbitrary real number.
Further, the point (1,-2) lies on the graph so that g(1) = -2.
Hence, 2a = -2 so that a = -1.
Thus, g(x) = -(x+1)/x.
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