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Mathematics 26 Online
Nickblade10:

Plz Help: What are the vertical asymptotes of the function f(x) = the quantity of 2x plus 8, all over x squared plus 5x plus 6? x = −3 and x = −2 x = −3 and x = 2 x = 1 and x = −2 x = 1 and x = 2

Nickblade10:

Vocaloid:

For vertical asymptotes, you want to find values where the function is undefined. In our case, the function is undefined when the denominator = 0. you can factor the denominator and set each factor equal to 0 to find the x values that will be your asymptotes.

Nickblade10:

i dont get it how do you do that?

Vocaloid:

Review factoring when you get a chance. x^2 + 5x + 6. Leading coefficient is 1, so factoring is simple. We pick two numbers that 1. Add up to the x coefficient, 5, and 2. Multiply to get the constant 6. 3 and 2 work. Therefore your factors are (x+3)(x+2) The denominator can now be written as (x+3)(x+2). The denominator is 0 when either x + 3 = 0 or x + 2 = 0 so you can solve these equations to get the two asymptotes.

Nickblade10:

Oh alright thank you

Nickblade10:

but thats not an answer :(

Nickblade10:

that gets you only positives which cant be

Vocaloid:

x + 3 = 0 Subtracting 3 from both sides gives us x = -3 Same logic for x + 2 = 0, subtract 2 to get x = -2

Nickblade10:

Oh i get it now. Thank you and im sorry for the stupidity

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