Please can someone help <span title="me:))))))))))))))))))))))))?">me:))))))))))))))))))))))...</span>

Question

ok, could you show me step by step on how to solve the point of discontinuity and vertical asympotoe of the following-

f(x)= x^2-2x-8/x+6

i really want to learn how to solve this but i can't figure it out.

please help

Answer 1

okay well we know that we cannot have 0 as the denominator in any fraction. in this fraction, the denominator is x+6.

so x+6 cannot = 0

solve it like a regular equation by subtracting 6 from both sides:

x cannot= -6

so at x=-6 that will be your vertical asymptote because the graph will never actually get there.

Answer 2

I simple terms the line

VERTICAL ASYMPTOTE

Ask yourself, plugging in what number(s) will give you an undefined value. zero in the deominator of any fraction is undefined. because you can&#039;t have pieces of nothing. therefore, if the problem were

EXAMPLE 1

4x/x^2 + 36

You would set the denominator to zero

x^2 +36 = 0

x^2 = -36. There is no such number, so you&#039;ll have no vertical asymptote

EXAMPLE 2

4x/x^2 - 36

x^2 -36 = 0

x^2 = 36

x = 6 and -6

remember (-6)(-6) = +36

Your vertical asymptotes would be 6 and -6

EXAMPLE 3

5/x-13

x- 13 = 0

x = 13

Your vertical asymptote would be 13 only

Answer 3

point of discontinuity is  at x=-6, because the denominator is zero at this point...

asymptote at x=-6

to find the asymptote, set the denominator to zero and solve for x

Answer 4

The point of discontinuity is where the function goes to H in a handbasket.

That happens whenever you divide by zero.

Since you have (x+6) in the denominator, you know that when x is -6, the denominator is zero.

Division by 0 is undefined.

The function is undefined when x is -6.  There is no continuity at that point.

As x approaches -6, the denominator gets smaller and smaller.  Thus, the function gets greater and greater as x gets closer to -6.  That is the vertical asymtote.