Respuesta :

Answer:

To find the vertical asympote or point of dicontinuities

Given function is,

[tex]f(x)=\frac{4x}{x^2-16}[/tex]

On solving the above equation, we get

[tex]f(x)=\frac{4x}{(x-4)(x+4)}[/tex]

f(x) is not defined when x=4 and x=-4 (that is when denominator tends to zero)

we get,

The point of discontinuity is at x=-4 and x=4

Vertical Asymptote: ertical Asymptote - when x approaches any constant value c, parallel to the y-axis, then the curve goes towards +infinity or – infinity.

we get that,

[tex]\lim _{x\to\pm4}f(x)=\lim _{x\to\pm4}\frac{4x}{x^2-16}=\pm\infty[/tex]

Vertical assymptote of f(x) is x=-4 and x=4

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