Find the horizontal asymptote of the given function. \[ h(x)=\frac{4-3 x}{-7 x+9} \] A. None B. $y=0$ C. $y=\frac{3}{7}$ D. $y=-\frac{3}{7}$

Step 1 ：Given the function \(h(x)=\frac{4-3 x}{-7 x+9}\)

Step 2 ：The horizontal asymptote of a rational function can be found by looking at the degrees of the numerator and denominator.

Step 3 ：If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y=0.

Step 4 ：If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.

Step 5 ：If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

Step 6 ：In this case, the degrees of the numerator and denominator are both 1, so the horizontal asymptote is the ratio of the leading coefficients.

Step 7 ：The leading coefficient of the numerator is -3 and the leading coefficient of the denominator is -7.

Step 8 ：So, the horizontal asymptote of the function is \(y=\frac{-3}{-7}\)

Step 9 ：Simplifying the above expression, we get \(y=\frac{3}{7}\)

Step 10 ：Final Answer: The horizontal asymptote of the function is \(\boxed{\frac{3}{7}}\)