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.