Question

practice assignment 5.6 rational functions
practice score: 2.85/5 answered: 3/5
question 4
let $f(x)=\frac{2x^{2}+7x - 15}{2x^{2}+7x + 6}$.
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 =

question help: video message instructor post to forum

Explanation:

Step1: Find y - intercept

Set \(x = 0\) in \(f(x)=\frac{2x^{2}+7x - 15}{2x^{2}+7x + 6}\). Then \(f(0)=\frac{2(0)^{2}+7(0)-15}{2(0)^{2}+7(0)+6}=\frac{- 15}{6}=-\frac{5}{2}\). So the y - intercept is at the point \((0,-\frac{5}{2})\).

Step2: Find x - intercepts

Set the numerator equal to zero: \(2x^{2}+7x - 15=0\). Factor the quadratic: \(2x^{2}+7x - 15=(2x - 3)(x + 5)=0\). Solving \((2x - 3)(x + 5)=0\) gives \(x=\frac{3}{2}\) or \(x=-5\). So the x - intercepts are at the points \((\frac{3}{2},0)\) and \((-5,0)\).

Step3: Find vertical asymptotes

Set the denominator equal to zero: \(2x^{2}+7x + 6 = 0\). Factor the quadratic: \(2x^{2}+7x + 6=(2x + 3)(x+2)=0\). Solving \((2x + 3)(x + 2)=0\) gives \(x=-\frac{3}{2}\) and \(x=-2\). So the vertical asymptotes are \(x =-\frac{3}{2}\) and \(x=-2\).

Step4: Find horizontal asymptote

Since the degree of the numerator and the denominator are the same (both are 2), the horizontal asymptote is \(y=\frac{a}{b}\), where \(a\) is the leading - coefficient of the numerator and \(b\) is the leading - coefficient of the denominator. Here \(a = 2\) and \(b = 2\), so \(y = 1\).

Answer:

1. \((0,-\frac{5}{2})\)
2. \((\frac{3}{2},0),(-5,0)\)
3. \(x=-\frac{3}{2},x=-2\)
4. \(y = 1\)