Question

rational functions worksheet 2
$f(x)=\frac{2x^{3}+4x^{2}-6x}{x^{2}+4x + 4}$
zeros
x - intercepts
y - intercepts
hole(s) and coordinate of hole(s)
vertical asymptote(s)
domain
horizontal/slant/quadratic asymptote
end behavior limits
behavior at vertical asymptotes

Explanation:

Step1: Factor the numerator and denominator

The numerator $2x^{3}+4x^{2}-6x = 2x(x^{2}+2x - 3)=2x(x + 3)(x - 1)$. The denominator $x^{2}+4x + 4=(x + 2)^{2}$. So $f(x)=\frac{2x(x + 3)(x - 1)}{(x + 2)^{2}}$.

Step2: Find the zeros

Set the numerator equal to 0: $2x(x + 3)(x - 1)=0$. Solving gives $x = 0,x=-3,x = 1$.

Step3: Find x - intercepts

The x - intercepts occur where $y = 0$, which is the same as the zeros, so $x=0,x=-3,x = 1$.

Step4: Find y - intercepts

Set $x = 0$ in the function: $f(0)=\frac{2\times0\times(0 + 3)\times(0 - 1)}{(0+2)^{2}}=0$.

Step5: Find holes

Set the common factors of the numerator and denominator equal to 0. Since there are no common non - constant factors, there are no holes.

Step6: Find vertical asymptotes

Set the denominator equal to 0: $(x + 2)^{2}=0$, so $x=-2$ is the vertical asymptote.

Step7: Find the domain

The domain is all real numbers except the value that makes the denominator 0. So the domain is $x
eq - 2$, or $(-\infty,-2)\cup(-2,\infty)$.

Step8: Find horizontal/slant/quadratic asymptote

Since the degree of the numerator ($n = 3$) is one more than the degree of the denominator ($m = 2$), there is a slant asymptote. Use long division: $\frac{2x^{3}+4x^{2}-6x}{x^{2}+4x + 4}=2x - 4+\frac{2x + 16}{x^{2}+4x + 4}$. The slant asymptote is $y = 2x-4$.

Step9: Find end - behavior limits

As $x\to\pm\infty$, $f(x)\to2x - 4$ (because the remainder $\frac{2x + 16}{x^{2}+4x + 4}\to0$ as $x\to\pm\infty$).

Step10: Find behavior at vertical asymptotes

As $x\to - 2^{+}$, $f(x)\to+\infty$ and as $x\to - 2^{-}$, $f(x)\to+\infty$ since $(x + 2)^{2}>0$ for $x
eq - 2$ and the sign of the numerator near $x=-2$ is non - zero.

Answer:

Zeros: $x = 0,x=-3,x = 1$
x - intercepts: $x = 0,x=-3,x = 1$
y - intercepts: $y = 0$
Hole(s) and coordinate of hole(s): None
Vertical Asymptote(s): $x=-2$
Domain: $(-\infty,-2)\cup(-2,\infty)$
Horizontal/Slant/Quadratic Asymptote: $y = 2x-4$
End Behavior Limits: As $x\to\pm\infty,f(x)\to2x - 4$
Behavior at Vertical Asymptotes: As $x\to - 2^{+},f(x)\to+\infty$; as $x\to - 2^{-},f(x)\to+\infty$