1.7 - 1.10 quiz ap precalc - no calculator. not eligible for reassessment\nname: \ndate: 12/3/25\nperiod…

Explanation:

Step1: Determine end - behavior of (f(x)=\frac{4x^3 + 1}{3x^2+2x + 3})

As (x\to\pm\infty), the end - behavior of a rational function is determined by the leading terms of the numerator and denominator. The degree of the numerator (n = 3) and the degree of the denominator (m=2). Since (n>m), as (x\to\infty), (f(x)\to\infty) and as (x\to-\infty), (f(x)\to-\infty).

Step2: Determine end - behavior of (h(x)=\frac{24}{x^2 - 9})

Factor the denominator (x^2 - 9=(x + 3)(x - 3)). As (x\to\pm\infty), the function (h(x)\to0) because the degree of the denominator ((m = 2)) is greater than the degree of the numerator ((n=0)).

Step3: Evaluate (\lim_{x\to-\infty}\frac{4x^3-7x^2}{2x^2 + 9})

Divide both the numerator and denominator by (x^2) (the highest - power of (x) in the denominator). We get (\lim_{x\to-\infty}\frac{4x-7}{2+\frac{9}{x^2}}). As (x\to-\infty), (\frac{9}{x^2}\to0), so (\lim_{x\to-\infty}\frac{4x-7}{2+\frac{9}{x^2}}=-\infty).

Step4: Find zeros of (f(x)=\frac{x^2-5x - 14}{x - 3})

Set the numerator equal to zero: (x^2-5x - 14=(x - 7)(x+2)=0). So (x = 7) and (x=-2) are the zeros of the function (since the denominator is non - zero at these points).

Step5: Find horizontal asymptote of (f(x))

For a rational function (\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}), if (n>m), there is no horizontal asymptote. If (n = m), the horizontal asymptote is (y=\frac{a_n}{b_m}), and if (n<m), the horizontal asymptote is (y = 0).

Answer:

End - behavior of (f(x)=\frac{4x^3 + 1}{3x^2+2x + 3}): as (x\to\infty), (f(x)\to\infty); as (x\to-\infty), (f(x)\to-\infty) End - behavior of (h(x)=\frac{24}{x^2 - 9}): as (x\to\pm\infty), (h(x)\to0) (\lim_{x\to-\infty}\frac{4x^3-7x^2}{2x^2 + 9}=-\infty) Zeros of (f(x)=\frac{x^2-5x - 14}{x - 3}): (x = 7) and (x=-2) Horizontal asymptote analysis depends on the rational function's numerator and denominator degrees as described above.