Answer:
To find the vertical asympote or point of dicontinuities
Given function is,

On solving the above equation, we get

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,

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