^{2}+ ...)/(-18x

^{2}- ...)

3x-6 = 0 x=2

f(x)= (6x-6)(x+3)/(-6x-2)(3x-6)

what are the vertical asymptotes?

what are the horizontal asymptotes?

what are the x-intercepts?

what are the y-intercepts?

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1. To find vertical asymptotes, set the denominator equal to zero, or:

(-6x-2)(3x-6) = 0

Now, we can solve for x,

remember if ab = 0, then either a=0 or b=0.

-6x-2 = 0; x=-1/3

3x-6 = 0; x=2

2. To find horizontal asymptotes, we compare leading terms. We know that the top and bottom are both polynomials with degree 2. Let's multiply them out to

(6x^{2} + ...)/(-18x^{2} - ...)

As x gets very large in either direction , f(x) tends towards the "a" term in these quadratics. Therefore, the horizontal asymptote is

y = 6/(-18); y=-1/3

Likewise, as x gets very small, f(x) tends towards the "c" term in these quadratics. Therefore, the other horizontal asymptote is

y=-18/12; y=-3/2

3. To find x-intercepts we set f(x) = 0 and solve for x. The first step is to find the numerator = 0, because

0/anything = 0

0 = (6x-6)(x+3)

another quadratic

6x-6 = 0 x=1

3x-6 = 0 x=2

3x-6 = 0 x=2

4. To find the y-intercepts, we set x=0 and solve for f(x)

(6(0)-6)(0+3)/(-6x-2)(3(0-6))

=(-6)(3)/((-2)(-6))

f(x) = 3/2

y= 3/2

f(x)= (6x-6)(x+3)/(-6x-2)(3x-6)=6(x-1)(x+3)/(-2)(3x+1)3(x-2)=-(x-1)(x+3)/(3x+1)(x-2)

The denominator is zero when x=-1/3, and also when x=2 both vertical asymptotes

for large x's we may recast f(x)=-(1-1/x)(1+3/x)/(3+1/x)(1-2/x) which goes to -1/3 as x→±∞

The function is zero for x=1, and x=-3. As for the y-intercept, when x=0 f(0)=-3/2

f(x) = -(x-1)(x+3)/((3x+1)(x-2))

No common factors in numerator and denominator so no holes.

Now multiply out and do long division:

f(x) = -(x^2+2x-3)/(3x^2-5x-2)

f(x) = -1/3 - (11x-7)/(3(3x+1)(x-2))

Horizontal asymptote at y = -1/3;

vertical asymptotes at x = -1/3 and x = 2.

x-intercepts when f(x) = 0 = (x-1)(x+3); i.e., x = -3, 1.

y-intercept when x = 0: y = -(-1)(3)/((1)(-2)) = -3/2

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