Find all vertical asymptotes and holes of each function. f(x)=x2−2x2x3−x2−6xf\left(x\right)=\frac{x^2-2x}{2x^3-x^2-6x}f(x)=2x3−x2−6xx2−2x

Hole(s): x = 0 x=0 x = 0 , x = 2 x=2 x = 2 , Vertical Asymptote(s): x = − 3 2 x=-\frac32 x = − 2 3

Find all vertical asymptotes and holes of each function. f ( x ) = x 2 − 2 x 2 x 3 − x 2 − 6 x f\left(x\right)=\frac{x^2-2x}{2x^3-x^2-6x} f ( x ) = 2 x 3 − x 2 − 6 x x 2 − 2 x

Hey, everyone. In this problem, we're asked to find all of the holes and vertical asymptotes of this function. And here I have the function f of x is equal to x squared minus 2x divided by 2x cubed minus x squared minus 6x. So let's go ahead and get started by factoring our numerator and our denominator. So first, taking a look at our numerator, it looks like I can pull a term of x out of each of these terms and factor that out, which leaves me with x times x minus 2. Now my numerator is fully factored. I can't factor that anymore. So let's go ahead and look at our denominator. Now in my denominator, it looks like I also have a common factor of x in all of these terms. So I can go ahead and factor that x out first, leaving with x times 2x squared minus x minus 6. Okay. Even though we already pulled that x out, we factored that. We can factor this even further by using the ac method and grouping. So first starting with the ac method, we want to go ahead and find something that multiplies to a times c, which in this case is negative 12 and adds to negative one. So with this in mind, I can actually have the factors negative 43, and I'm gonna go ahead and separate this out and expand it so that I can then use grouping. So this expands out into 2x squared minus 4x plus 3x minus 6. And now I can go ahead and group these terms in order to factor this. So I'm going to group my first two terms and my last two terms and go ahead and pull out any common factors. So for 2x squared and 4x, I can go ahead and pull out a 2x of both of these terms. This leaves me with x minus 2 inside of my parentheses there. Then over here, I can pull a 3 out of both of these terms. So that leaves me with 3 times x minus 2. Now I can go ahead and see that these are common factors. So this is fully factored out into 2x+3 timesxminus2. So taking this all the way back to our original fraction, let's go ahead and rewrite our function yet again. So my function is now fully factored into x times x minus 2 divided by x. And then my factored function here, 2x+3 timesxminus2. Okay. Now that we have fully factored that, no factoring left to do here, let's go ahead and take a look at our common factors. Well, it looks like I have 2 common factors. I have x minus 2 in both my numerator and my denominator, but I also have an x in both of them. So that tells me that I need to take both of these and set them equal to 0 to find the holes, which I'm going to do over here. So I'm going to take x, my first common factor, and set it equal to 0 and then my second common factor, x minus 2, and set that equal to 0 as well. Now, x equals 0 is already done, so I know I have a hole at x equals 0. But then I want to add 2 to both sides to solve for x here, leaving me with x is equal to 2. So that means I have a second hole at x equals 2. It's possible to have no holes or more than 1 or just one holes in your rational function. So let's go ahead and move on to our vertical asymptotes. Now since we already took care of those common factors with our holes, we can go ahead and cancel them out here, And that leaves me with just 1 over 2x+3 for my function in lowest terms, a much more simple looking function than what I had before. So from here, I can just set my denominator equal to 0, which means I can subtract 3 from both sides, leaving me with 2x is equal to negative 3. Then dividing both sides by 2 leaves me finally with x is equal to negative 3 over 2 as my vertical asymptote. So I have one vertical asymptote at negative 3 over 2 and I don't have to worry about a second one I'm done here. That was a longer one but we identified all of the holes and vertical asymptotes of the graph of this rational function. I'll see you in the next one.

Find all vertical asymptotes and holes of each function. f ( x ) = x 2 − 2 x 2 x 3 − x 2 − 6 x f\left(x\right)=\frac{x^2-2x}{2x^3-x^2-6x} f ( x ) = 2 x 3 − x 2 − 6 x x 2 − 2 x

Hole(s): x = 0 x=0 x = 0 , x = 2 x=2 x = 2 , Vertical Asymptote(s): x = − 3 2 x=-\frac32 x = − 2 3

Hole(s): None, Vertical Asymptote(s): x = 0 , x = 2 , x = − 3 2 x=0,x=2,x=-\frac32 x = 0 , x = 2 , x = − 2 3

Hole(s): x = 0 x=0 x = 0 , Vertical Asymptote(s): x = − 3 2 x=-\frac32 x = − 2 3

Hole(s): x = 0 x=0 x = 0 , x = 2 x=2 x = 2 , Vertical Asymptote(s): x = 3 2 x=\frac32 x = 2 3

5. Rational Functions

Asymptotes

College Algebra

5. Rational Functions

Asymptotes

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Multiple Choice

Find all vertical asymptotes and holes of each function.
$f\left(x\right)=\frac{x^2-2x}{2x^3-x^2-6x}$

Hole(s): None, Vertical Asymptote(s): $x=0,x=2,x=-\frac32$

Hole(s): $x=0$ , Vertical Asymptote(s): $x=-\frac32$

Hole(s): $x=0$ , $x=2$ , Vertical Asymptote(s): $x=-\frac32$

Hole(s): $x=0$ , $x=2$ , Vertical Asymptote(s): $x=\frac32$

Verified step by step guidance

Start by factoring the numerator and the denominator of the function f(x) = \frac{x^2 - 2x}{2x^3 - x^2 - 6x}.

Factor the numerator: x^2 - 2x can be factored as x(x - 2).

Factor the denominator: 2x^3 - x^2 - 6x can be factored by first taking out the common factor x, giving x(2x^2 - x - 6). Then, factor the quadratic 2x^2 - x - 6 further.

Identify the common factors between the numerator and the denominator. These common factors will indicate the location of holes in the graph of the function.

Determine the values of x that make the denominator zero but are not canceled by the numerator. These values are the vertical asymptotes of the function.

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