x ≠ 3

x ≠ -1

x ≠ 3, -1

x ≠ 3, 1

(1, 0)

(-1, 0)

(1, 0), (-1, 0)

(0, 0), (1, 0)

The horizontal asymptote is at y = 0.

There is a hole at (6, ¾ ).

The vertical asymptote is at x = -2.

The x and y intercepts are at the origin.

The domain is all real numbers except -2.

Crossing the x-axis

An absence of dirt.

Crossing a vertical asymptote.

A factor that cancels out.

(-1, -2)

(-1, ½)

(-1, -½)

there isn't one

V.A. at x = 1

No Holes

H.A. at y = 0

y-intercept at (0, 1/2)

V.A. at x = 2

No Holes

H.A. at y = -1/3

y-intercepts at (0, 1/2)

Yes

No

Cannot be determined

y = x

y = x - 8

y = x - 4

y = x + 4

y = 3x - 12

y = x + 5

y = 3x - 5

y = 3x - 7

(0, 1)

[-1, 0] u [1, oo)

[0,1]

(-1, 0) u (1, oo)

(-oo, -4) u (3, oo)

(-4, 3)

[-4, 3]

(-oo, -4] u [3, oo)

$f(x)=\frac{x^2}{x^2+x-12}$

$f(x)=\frac{4x^2}{x^2-x-12}$

$f(x)=\frac{2x^2-2}{2x^2}$

$f(x)=\frac{x}{x^2+x-12}$

True, each vertical asymptote is obtained by setting the factors of the denominator = 0.

True, each vertical asymptote is obtained by setting the factors of the numerator = 0.

False, there can be only one determined by the power of the numerator compared to the power of the denominator.

True, a graph can cross a horizontal asymptote because it really dictates the end behavior.

False.

 $Domain:\ x\ne\pm3\ \ \ \ Range:\ \infty$ 

 $Domain:\ \left(-\infty,\ -3\right)\cup\left(-3,\ 3\right)\cup\left(3,\ +\infty\right)\ \ \ \$   $Range:\ \left(-\infty,\ 0\right)\cup\left(0,\ +\infty\right)$  

 $Domain:\ \left(-\infty,\ -3\right)\cup\left(-3,\ 3\right)\cup\left(3,\ +\infty\right)\ \ \ \$  
 $Range:\ \left(-\infty,\ +\infty\right)$  

$f\left(x\right)=\frac{2x^2-2x+24}{x^2-x-12}$

$h\left(x\right)=\frac{-2x^2-2x-24}{x^3+x^2-12}$

$g\left(x\right)=\frac{2\left(x-4\right)\left(x+3\right)}{\left(x+4\right)\left(x+3\right)}$

$f\left(x\right)=\frac{-2x^2+2x+24}{x^3-x^2-12x}$