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Find any points of discontinuity for the rational function. y = (x - 1)/(x ^ 2 - 2x - 8)

Find Any Points Of Discontinuity For The Rational Function Y X 1x 2 2x 8 class=

Sagot :

jacob193

Answer:

Asymptotic discontinuities at [tex]x = (-2)[/tex] and [tex]x = 4[/tex].

Step-by-step explanation:

A linear function has an asymptotic discontinuity at [tex]x = a[/tex] if [tex](x - a)[/tex] is a factor of the denominator after simplification.

The numerator of this function, [tex](x - 1)[/tex], is linear in [tex]x[/tex].

The denominator of this function, [tex](x^{2} - 2\, x - 8)[/tex], is quadratic in [tex]x[/tex]. Using the quadratic formula or otherwise, factor the denominator into binominals:

[tex]\begin{aligned}y &= \frac{(x - 1)}{x^{2} - 2\, x - 8} \\ &= \frac{(x - 1)}{(x - 4)\, (x + 2)}\end{aligned}[/tex].

Simplify the function by liminating binomials that are in both the numerator and the denominator.

Notice that in the simplified expression, binomial factors of the denominator are [tex](x - 4)[/tex] and [tex](x + 2)[/tex] (which is equivalent to [tex](x - (-2))[/tex].) Therefore, the points of discontinuity of this function would be [tex]x = 4[/tex] and [tex]x = (-2)[/tex].

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