Question

1. identify any horizontal asymptotes, vertical asymptotes, and holes of the function below: (5 pts)$g(x) = \frac{x^2 - x - 6}{x^3 - 8x^2 + 15x}$a) horizontal asymptote(s): ________b) vertical asymptote(s): ______c) hole(s): ______d) zero(s): ________10) using arrows and limit notation, determine the end behavior of the following polynomials: (5 pts)a) $p(x) = 5x^3 + 2x - 7$b) $p(x) = 6x^5 - 4x^6 + 9x$11)given that $f(x) = x^2 + 3x$ and $g(x) = - 2x - 1$, calculate the following: (5 pts)a) $f(g(-1))$b) $g(f(-1))$

Explanation:

Problem 9

Step1: Simplify the rational function

First, factor numerator and denominator:
Numerator: $x^2 - x - 6 = (x-3)(x+2)$
Denominator: $x^3 - 8x^2 + 15x = x(x^2-8x+15) = x(x-3)(x-5)$
Cancel common factor $(x-3)$ (for $x
eq3$):
$g(x)=\frac{(x+2)}{x(x-5)}$, $x
eq3$

Step2: Find horizontal asymptote

Compare degrees: numerator degree=1, denominator degree=3. Since $1<3$, horizontal asymptote is $y=0$.

Step3: Find vertical asymptotes

Set simplified denominator to 0: $x(x-5)=0 \implies x=0, x=5$

Step4: Find holes

Holes occur at canceled factors: $x-3=0 \implies x=3$

Step5: Find zeros

Set simplified numerator to 0: $x+2=0 \implies x=-2$

Step1: Analyze end behavior for a)

For $p(x)=5x^3+2x-7$, leading term is $5x^3$ (odd degree, positive leading coefficient):
As $x\to+\infty$, $p(x)\to+\infty$; As $x\to-\infty$, $p(x)\to-\infty$

Step2: Analyze end behavior for b)

For $p(x)=6x^5-4x^6+9x$, rewrite in standard form: $p(x)=-4x^6+6x^5+9x$ (even degree, negative leading coefficient):
As $x\to+\infty$, $p(x)\to-\infty$; As $x\to-\infty$, $p(x)\to-\infty$

Step1: Solve part a) $f(g(-1))$

First calculate $g(-1)$:
$g(-1)=-2(-1)-1=2-1=1$
Then substitute into $f(x)$:
$f(1)=(1)^2+3(1)=1+3=4$

Step2: Solve part b) $g(f(-1))$

First calculate $f(-1)$:
$f(-1)=(-1)^2+3(-1)=1-3=-2$
Then substitute into $g(x)$:
$g(-2)=-2(-2)-1=4-1=3$

Answer:

a) $y=0$
b) $x=0$, $x=5$
c) $x=3$
d) $x=-2$

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Problem 10