finding vertical tangent lines and reasoning

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October 15th 2009, 09:30 AM
hazecraze

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?
October 15th 2009, 09:37 AM
Mush

Quote:

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}
October 15th 2009, 10:26 AM
hazecraze

So the denominator would have to be a root function or a fraction to be a vertical tangent line?
October 15th 2009, 10:42 AM
Mush

Quote:

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.