Question

assignment 5.6 rational functions
score: 10.5/18 answered: 10/18
question 13
let $f(x)=\frac{2x^{2}-9x + 9}{3x^{2}-1x - 2}$.
this function has:

1. a y - intercept at the point
2. x - intercepts at the point(s)
3. vertical asymptotes at $x=$
4. horizontal asymptote at $y=$

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Explanation:

Step1: Find y - intercept

Set \(x = 0\) in \(f(x)=\frac{2x^{2}-9x + 9}{3x^{2}-x - 2}\). Then \(f(0)=\frac{2(0)^{2}-9(0)+9}{3(0)^{2}-0 - 2}=-\frac{9}{2}\). So the y - intercept is at the point \((0,-\frac{9}{2})\).

Step2: Find x - intercepts

Set \(y = f(x)=0\), so \(2x^{2}-9x + 9 = 0\). Factor the quadratic: \(2x^{2}-9x + 9=(2x - 3)(x - 3)=0\). Solving \((2x - 3)(x - 3)=0\) gives \(x=\frac{3}{2}\) and \(x = 3\). The x - intercepts are at the points \((\frac{3}{2},0)\) and \((3,0)\).

Step3: Find vertical asymptotes

Set the denominator \(3x^{2}-x - 2=0\). Factor the quadratic: \(3x^{2}-x - 2=(3x + 2)(x - 1)=0\). Solving \((3x + 2)(x - 1)=0\) gives \(x=-\frac{2}{3}\) and \(x = 1\). So the vertical asymptotes are \(x=-\frac{2}{3}\) and \(x = 1\).

Step4: Find horizontal asymptote

Since the degree of the numerator \(n = 2\) and the degree of the denominator \(m = 2\) and the leading - coefficient of the numerator \(a=2\) and the leading - coefficient of the denominator \(b = 3\), the horizontal asymptote is \(y=\frac{a}{b}=\frac{2}{3}\).

Answer:

1. \((0,-\frac{9}{2})\)
2. \((\frac{3}{2},0),(3,0)\)
3. \(x=-\frac{2}{3},x = 1\)
4. \(y=\frac{2}{3}\)