Show Intro/Instructions Determine the horizontal asymptotes of the following functions. Graph the function if you need to check your answer. a. The function $f(x)=\frac{17+9 x}{3 x-5}$ has a horizontal asymptote at... \[ y= \] Preview b. The function $f(x)=\frac{2 x^{2}+4 x-10}{2 x^{2}-x-13}$ has a horizontal asymptote at... \[ y= \] Preview c. The function $f(x)=\frac{x+4000}{x^{2}-7}$ has a horizontal asymptote at... \[ y= \] Preview d. The function $f(x)=\frac{5 x+1}{6-15 x}$ has a horizontal asymptote at... \[ y= \] Preview

Step 1 ：The horizontal asymptote of a function can be determined by looking at the degrees of the polynomials in the numerator and the denominator.

Step 2 ：If the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, the horizontal asymptote is at y = 0.

Step 3 ：If the degrees are the same, the horizontal asymptote is at y = the ratio of the leading coefficients.

Step 4 ：If the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator, there is no horizontal asymptote.

Step 5 ：Let's apply this to the first function \(f(x)=\frac{17+9 x}{3 x-5}\).

Step 6 ：The degree of the polynomial in the numerator is 1 (from the term 9x) and the degree of the polynomial in the denominator is also 1 (from the term 3x). Therefore, the horizontal asymptote is at y = the ratio of the leading coefficients, which is \(\frac{9}{3}\).

Step 7 ：Final Answer: The function \(f(x)=\frac{17+9 x}{3 x-5}\) has a horizontal asymptote at \(y=\boxed{3}\).