Find all vertical asymptotes and holes of each function. f(x)=−5x(2x−3)2f\left(x\right)=\frac{-5x}{\left(2x-3\right)^2}f(x)=(2x−3)2−5x

Hole(s): None , Vertical Asymptote(s): x=32x=\frac32x=23

Find all vertical asymptotes and holes of each function. f ( x ) = − 5 x ( 2 x − 3 ) 2 f\left(x\right)=\frac{-5x}{\left(2x-3\right)^2} f ( x ) = ( 2 x − 3 ) 2 − 5 x

In this problem, I'm asked to find all of the holes and vertical asymptotes of the function. And here, I have the function f of x is equal to negative 5x divided by 2x minus 3 squared. Now, remember, we start finding holes in vertical asymptotes the same way by factoring our function, so let's go ahead and do that. Well, my numerator cannot be further factored. It's just negative 5x, nothing I can factor there. And then my denominator is 2x minus 3 squared, which is also already fully factored. So I actually can't factor anything here. And looking at this, I also have no common factors that are going to cancel out. Now that tells me that I have no holes in my function, which is possible. You won't always have a hole in your function because you won't always have a common factor that needs to get canceled. Okay. Now that we know that there are no holes in our function, we can proceed with finding our vertical asymptotes by setting our denominator equal to 0. Now when I do this, if I cancel out this squared by taking the square root, I know that I'm just gonna be left with 0 on that side. So this is really just 2x minus 3 equals 0. Now, in order to solve for x, I want to add 3 to both sides, canceling 3 over here, leaving me with 2x is equal to 3. Finally, to isolate x, I want to divide both sides by 2, leaving me with x is equal to 3 halves. So I have one vertical asymptote at x equals threetwo. And that's my only vertical asymptote. So that's all I have for this problem. I'll see you in the next one.

Find all vertical asymptotes and holes of each function. f(x)=−5x(2x−3)2f\left(x\right)=\frac{-5x}{\left(2x-3\$$right)^2$$}f(x)=(2x−3)2−5x

Hole(s): None , Vertical Asymptote(s): x=32x=\frac32x=23

Hole(s): x=0x=0x=0 , Vertical Asymptote(s): x=32x=\frac32x=23

Hole(s): x=32x=\frac32x=23 , Vertical Asymptote(s): x=32x=\frac32x=23

Hole(s): x=0x=0x=0 , Vertical Asymptote(s): x=0x=0x=0

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{-5x}{\left(2x-3\right)^2}$$

Hole(s): $x=0$ , Vertical Asymptote(s): $x=$\frac$32$

Hole(s): $x=$\frac$32$ , Vertical Asymptote(s): $x=$\frac$32$

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

Hole(s): None , Vertical Asymptote(s): $x=$\frac$32$

검증된 단계별 안내

Identify the function given: $ f(x) = \frac{-5x}{(2x-3)^2} $. This is a rational function, which means it can have vertical asymptotes and holes.

To find vertical asymptotes, set the denominator equal to zero and solve for $ x $. The denominator is $(2x-3)^2$, so set $2x-3 = 0$.

Solve the equation $2x-3 = 0$ to find the value of $x$ that makes the denominator zero. This will give you the vertical asymptote.

To check for holes, look for common factors in the numerator and the denominator. The numerator is $-5x$ and the denominator is $(2x-3)^2$. There are no common factors, so there are no holes.

Conclude that the function has a vertical asymptote at $x = \frac{3}{2}$ and no holes.

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Struggling with College Algebra?

Find all vertical asymptotes and holes of each function. f(x)=−5x(2x−3)2f\(\left\)(x\(\right\))=\(\frac{-5x}{\left(2x-3\right)^2}\)f(x)=(2x−3)2−5x

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Asymptotes practice set

Find all vertical asymptotes and holes of each function.
$f\(\left\)(x\(\right\))=\(\frac{-5x}{\left(2x-3\right)^2}\)$