# Thread: finding vertical tangent lines and reasoning

1. ## finding vertical tangent lines and reasoning

Find vertical tangent lines of $y=(2x-6)/(sqrt x^2+3)$. (how do you do square root symbols?)

$y'=[(12-12x)/(2sqrtx^2+)]/(x^2+3)$

$(x^2+3)=0$
$x=+or- (-sqrt3)$, x=complex number, so there are no vertical tangent lines?

Edit: Or actually, $(x^2+3)$ will never be zero, so the denominator would have to be a root function or a fraction to be a vertical tangent line?

2. Originally Posted by hazecraze
Find vertical tangent lines of $y=(2x-6)/(sqrt x^2+3)$. (how do you do square root symbols?)

$y'=[(12-12x)/(2sqrtx^2+)]/(x^2+3)$

$(x^2+3)=0$
$x=+or- (-sqrt3)$, x=complex number, so there are no vertical tangent lines?

Edit: Or actually, $(x^2+3)$ will never be zero, so the denominator would have to be a root function or a fraction to be a vertical tangent line?
There are no verticals, you are right.

To do square roots, put backslash before the word sqrt.

Also, to do a fraction, write \frac{numerator}{denominator}

3. So the denominator would have to be a root function or a fraction to be a vertical tangent line?

4. Originally Posted by hazecraze
So the denominator would have to be a root function or a fraction to be a vertical tangent line?
In order for a function to have vertical asymptotes, there must be points at which the function is not defined.

In functions like these, the most common 'cause' of a function being undefined is a divide by zero at for some value of x. Thus, yes, the denominator of such a function must be equal to zero at some value of x.

There are other reasons why a function may be undefined, but in this type of example, you'd need some sort of division by 0, and that does not occur in the real plane for this function.