Answer :
points of discontinuity are x = -4&x = +4.
x = +4 is removable discontinuity
The x- intercept is x = 0The y- intercept is y= -(1/16).
How to find the points of discontinuity of the rational function & find the x - and y -intercepts of the rational function?
points of discontinuity:
In rational functions, points of discontinuity are defined as fractions with undefined or zero as their denominator. When the denominator of a fraction is 0, the fraction is no longer defined and shows up as a hole or break in the graph.
points of discontinuity removable:
A point on the graph that is removable discontinuity is one that may be filled in with a single value despite not existing.If an is zero for a factor in the denominator that shares a factor with a factor in the numerator, a removable discontinuity at x=a in the graph of a rational function occurs.
points of discontinuity non removable:
A function's break that cannot be repaired with a single point is referred to as a non-removable discontinuity. It cannot be filled in with a single point, a vertical asymptote (a vertical line that the graph approaches but never crosses because the function is undefined at that x-value) is non removable.
given that y=1/x²-16
write 16 as a 4²
[tex]y = \frac{1}{ {x}^{2} - {4}^{2} } [/tex]
[tex]y = \frac{1}{(x + 4)(x - 4)} [/tex]
now,
domain : x !=4.
points of discontinuity are
(x+4) = 0 (x-4) = 0
x = -4 x = +4
Here x = +4 is removable discontinuity.
The x- intercept is
if x = 0, then
[tex]y = \frac{ 1}{ - 16} [/tex]
The y- intercept is
if y = 0
x = 0.
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