26 Section 5.5 Key Features of Rational Functions
a. [latex]f(x)=\frac{2x^4(x+3)^2(x-12)(x+5)}{(x+5)(x+12)(x^2-4)}[/latex]
b. [latex]g(x)=\frac{(x-1)(x^2+1)}{(2x-1)(x+1)(x+4)}[/latex]
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a. [latex]f(x)[/latex] has the following characteristics:
- Vertical asymptotes at [latex]x=-12[/latex], [latex]x=2[/latex], and [latex]x=-2[/latex].
- Hole at [latex]x=-5[/latex].
- No horizontal asymptote since the numerator’s degree is greater than the denominator’s degree.
- Horizontal intercepts at [latex](0,0)[/latex], [latex](-3,0)[/latex] and [latex](12,0)[/latex].
- Vertical intercept at [latex](0,0)[/latex].
- Domain is [latex](-\infty,-12)\cup (-12,-5)\cup (-5,-2)\cup (-2,2)\cup (2,\infty)[/latex].
b. [latex]g(x)[/latex] has the following characteristics:
- Vertical asymptotes at [latex]x=\frac{1}{2}[/latex], [latex]x=-1[/latex], and [latex]x=-4[/latex].
- No holes.
- Horizontal asymptote at [latex]y=\frac{1}{2}[/latex].
- Horizontal intercept at [latex](1,0)[/latex].
- Vertical intercept at [latex](0,\frac{1}{4})[/latex].
- Domain is [latex](-\infty,-4)\cup (-4,-1)\cup (-1,\frac{1}{2})\cup (\frac{1}{2},\infty)[/latex].
2. Create a formula for a rational function with the given criteria.
A rational function, [latex]m[/latex], has a vertical asymptote at [latex]x=-4[/latex], holes at [latex]x=1[/latex] and [latex]x=3[/latex], and a horizontal asymptote at [latex]y=\frac{5}{4}[/latex]. There is also one horizontal intercept [latex](-6,0)[/latex]. Both the numerator and denominator have degree 5.
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One formula that fits this criteria is: [latex]m(x)=\frac{5(x-1)(x+6)(x-3)^3}{4(x-1)^2(x-3)(x+4)^2}[/latex].
To create this equation, start with the factor that creates the vertical asymptote of [latex](x+4)[/latex] in the denominator. Then also create factors for the holes that go in the numerator and denominator [latex](x-1)(x-3)[/latex]. So far, this leads to:
[latex]m(x)=\frac{(x-1)(x-3)}{(x-1)(x-3)(x+4)}[/latex]
To factor in the horizontal intercept, put the factor [latex](x+6)[/latex] into the numerator.
[latex]m(x)=\frac{(x-1)(x-3)(x+6)}{(x-1)(x-3)(x+4)}[/latex]
There are two more things to account for; first, the horizontal asymptote at [latex]y=\frac{5}{4}[/latex].
[latex]m(x)=\frac{5(x-1)(x-3)(x+6)}{4(x-1)(x-3)(x+4)}[/latex]
Then finally, the numerator and denominator are only degree 3, so add some multiplicity to reach degree 5.
[latex]m(x)=\frac{5(x-1)(x+6)(x-3)^3}{4(x-1)^2(x-3)(x+4)^2}[/latex]