How do you find the vertical asymptotes #(x^2+4)/(6x-5x^2)#?

Question

Serenity Jones · Accepted Answer

The answer would be got by equating denominator to zero and solving. Here it would be #x=0# or #x=6/5#

The given function.

#(x^2+4)/(6x-5x^2)#

To find the Vertical Asymptotes factor both numerator and denominator first.

The numerator is not factorable.

The denominator can be factored as # x(6-5x)#

The next step is to cancel the common factors if any from the numerator and denominator. In our problem there is none. This is done to eliminate holes

Finally, equate the factors remaining in the denominator to zero and solve.

#x(6-5x) =0#

#x=0 or 6-5x =0 # zero product rule

#x= 0 or x=6/5 # These are the equations of the Vertical Asymptotes.

Serenity Jones · Answer

undefined To find the vertical asymptotes of the function $ \frac{x^2 + 4}{6x - 5x^2} $, set the denominator equal to zero and solve for $ x $. The values of $ x $ obtained from this process will represent the vertical asymptotes of the function.

Lincoln Anderson · Answer

undefined To find the vertical asymptotes of $ \frac{x^2 + 4}{6x - 5x^2} $, set the denominator equal to zero and solve for $ x $.

$6x - 5x^2 = 0$

Factor out $x$: $x(6 - 5x) = 0$

Set each factor equal to zero: 
1. $x = 0$ 
2. $6 - 5x = 0$ 
   
Solve for $x$ in the second equation:
$$6 - 5x = 0$$
$$6 = 5x$$
$$x = \frac{6}{5}$$

So, the vertical asymptotes are $x = 0$ and $x = \frac{6}{5}$.