Find all vertical asymptotes and holes of each function. f(x)=x2+10x+252x2+8x−10f\left(x\right)=\frac{x^2+10x+25}{2x^2+8x-10}f(x)=2x2+8x−10x2+10x+25

Hole(s): x = − 5 x=-5 x = − 5 , Vertical Asymptote(s): x = 1 x=1 x = 1

Find all vertical asymptotes and holes of each function. f ( x ) = x 2 + 10 x + 25 2 x 2 + 8 x − 10 f\left(x\right)=\frac{x^2+10x+25}{2x^2+8x-10} f ( x ) = 2 x 2 + 8 x − 10 x 2 + 10 x + 25

In this problem, we want to find all of the holes and vertical asymptotes of the function. And here, I have the function f of x is equal to x squared plus 10x plus 25 over 2x squared plus 8x minus 10. So let's go ahead and factor both the numerator and the denominator here. Now, in my numerator, I see that this is actually a perfect square trinomial, so this will factor into x+5 squared. Now, looking at my denominator here in all of these terms, it looks like I can pull out a greatest common factor of 2 from each of these. So pulling out that 2 leaves me with 2 times x squared plus 4x minus 5. Now this can be further factored here using the ac method looking for something that multiplies to negative 5 and adds to 4. So this denominator will further factor into 2, leaving that 2 there, times x minus 1 times x plus 5. Then my numerator pulling that over here is still x+5 squared, and now we're fully factored. So here, it looks like I do have a common factor of x plus 5. So I can go ahead and take that common factor, x plus 5, and set it equal to 0 in order to find the hole in the graph of my rational function. So solving for x here, if I subtract 5 from both sides, I'm left with x is equal to negative 5. And that's where the only hole in my graph is, x equals negative 5. I don't have to worry about anything else because I only have that one common factor. Now, going back to our function and canceling out that common factor, I'm left with x+5 divided by 2 times x minus 1 as my function in lowest terms. Now that we're in lowest terms, we can go ahead and take our denominator and set it equal to 0 in order to find our vertical asymptote or asymptotes if that works out that way. Now here's solving for x. If I divide both sides by 2, I know that that's not going to do anything since I just have 0. So this is really just x minus 1 equals 0, which, isolating x here, I can just add 1 to both sides, leaving me with x is equal to 1 as my vertical asymptote. Now, I only have one vertical asymptote here to go with the one hole that I have. And we're done with this problem. I'll see you in the next Video.

Find all vertical asymptotes and holes of each function. f ( x ) = x 2 + 10 x + 25 2 x 2 + 8 x − 10 f\left(x\right)=\frac{x^2+10x+25}{2x^2+8x-10} f ( x ) = 2 x 2 + 8 x − 10 x 2 + 10 x + 25

Hole(s): x = − 5 x=-5 x = − 5 , Vertical Asymptote(s): x = 1 x=1 x = 1

Hole(s): None, Vertical Asymptote(s): x = − 5 , x=-5, x = − 5 , x = 1 x=1 x = 1

Hole(s): x = 1 x=1 x = 1 , Vertical Asymptote(s): x = − 5 x=-5 x = − 5

Hole(s): x = − 5 x=-5 x = − 5 , Vertical Asymptote(s): x = − 1 x=-1 x = − 1

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.
$f\left(x\right)=\frac{x^2+10x+25}{2x^2+8x-10}$

Hole(s): None, Vertical Asymptote(s): $x=-5,$ $x=1$

Hole(s): $x=-5$ , Vertical Asymptote(s): $x=1$

Hole(s): $x=1$ , Vertical Asymptote(s): $x=-5$

Hole(s): $x=-5$ , Vertical Asymptote(s): $x=-1$

Verified step by step guidance

Factor the numerator and the denominator of the function f(x) = \frac{x^2 + 10x + 25}{2x^2 + 8x - 10}.

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

Set the common factors equal to zero to find the x-values where the holes occur.

For vertical asymptotes, set the denominator equal to zero and solve for x. These values are where the function is undefined and vertical asymptotes occur.

Verify the solutions by checking the simplified form of the function and confirming the locations of holes and vertical asymptotes.

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