Question

1. graph and analyze h(x) = (x^2 + 2x - 15)/(x^2 - 1) without using a graphing calculator.

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
crosses ha? where?
*go by 2s on y axis

Explanation:

Step1: Factor the numerator and denominator

Factor $x^{2}+2x - 15=(x + 5)(x-3)$ and $x^{2}-1=(x + 1)(x - 1)$. So $h(x)=\frac{(x + 5)(x - 3)}{(x + 1)(x - 1)}$.

Step2: Find the zeros

Set the numerator equal to zero: $(x + 5)(x - 3)=0$. Then $x=-5$ or $x = 3$.

Step3: Find the x - intercepts

The x - intercepts occur where $y = 0$, which is the same as the zeros. So x - intercepts are $x=-5$ and $x = 3$.

Step4: Find the y - intercepts

Set $x = 0$. Then $h(0)=\frac{0^{2}+2\times0-15}{0^{2}-1}=\frac{-15}{-1}=15$.

Step5: Find holes

Since there are no common factors in the numerator and denominator after factoring, there are no holes.

Step6: Find vertical asymptotes

Set the denominator equal to zero: $(x + 1)(x - 1)=0$. So $x=-1$ and $x = 1$ are the vertical asymptotes.

Step7: Find the domain

The domain is all real numbers except the values that make the denominator zero. So the domain is $x
eq-1,x
eq1$, or $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$.

Step8: Find horizontal/slant/quadratic asymptote

Since the degree of the numerator and denominator are the same (both degree 2), the horizontal asymptote is $y=\frac{1}{1}=1$ (the ratio of the leading - coefficients).

Step9: Find end - behavior limits

$\lim_{x
ightarrow\pm\infty}h(x)=1$ because of the horizontal asymptote.

Step10: Behavior at vertical asymptotes

$\lim_{x
ightarrow - 1^{-}}h(x)=\infty$, $\lim_{x
ightarrow - 1^{+}}h(x)=-\infty$, $\lim_{x
ightarrow1^{-}}h(x)=-\infty$, $\lim_{x
ightarrow1^{+}}h(x)=\infty$.

Step11: Check if it crosses the horizontal asymptote

Set $h(x)=1$. Then $\frac{x^{2}+2x-15}{x^{2}-1}=1$. Cross - multiply: $x^{2}+2x-15=x^{2}-1$. Subtract $x^{2}$ from both sides: $2x-15=-1$, $2x = 14$, $x = 7$.

Answer:

Property	Value
X - Intercepts	$x=-5,x = 3$
Y - Intercepts	$y = 15$
Hole(s) and coordinate of hole(s)	None
Vertical Asymptote(s)	$x=-1,x = 1$
Domain	$(-\infty,-1)\cup(-1,1)\cup(1,\infty)$
Horizontal/Slant/Quadratic Asymptote	$y = 1$

| End Behavior Limits | $\lim_{x
ightarrow\pm\infty}h(x)=1$ |
| Behavior at Vertical Asymptotes | $\lim_{x
ightarrow - 1^{-}}h(x)=\infty$, $\lim_{x
ightarrow - 1^{+}}h(x)=-\infty$, $\lim_{x
ightarrow1^{-}}h(x)=-\infty$, $\lim_{x
ightarrow1^{+}}h(x)=\infty$ |

Crosses HA? Where?

Yes, at $x = 7$