A function f has a horizontal asymptote of y=-4, a vertical asymptote of x=3 and an x-intercept at (1, 0).

(a) We use the vertical asymptote, x=3, to help us solve for c. We let x be 3. We need to find c so that the denominator, x+c, is 0, and the fraction is undefined.

$x+c=0\\3+c=0\\c=-3.$ 

We let f(x) equal to y and plug in the value of c. We then have

$y=\frac{ax+b}{x-3}\\y(x-3)=\frac{ax+b}{x-3}(x-3)\\y(x-3)=ax+b\\xy-3y=ax+b\\xy-ax=b+3y\\x(y-a)=b+3y\\x=\frac{3y+b}{-a+y}.$ 

Now we use the horizontal asymptote y=-4 to help us solve for a. Using similar logic as we solved for c to solve for a, we get

 $-a+y=0\\-a-4=0\\a=-4.$

We then plug the values of a and c into $f(x)=\frac{ax+b}{x-3}$ to get $f(x)=\frac{-4x+b}{x-3}$. We plug in the point (1, 0) into the equation to solve for b. $f(x)=\frac{-4x+b}{x-3}\\0=\frac{-4(1)+b}{1-3}\\0=\frac{-4+b}{-2}\\0=-4+b\\b=4.$

Therefore $f(x) = \frac{ax+b}{x+c}$ is $\boxed{f(x) = \frac{-4x+4}{x-3}.}$ 

Use similar logic to solve for part (b).