Question

use the denominator to identify all places where the functions below are discontinuous. then check the value of the numerator at each discontinuity to determine if each discontinuity is a removable discontinuity (\hole\) or an infinite discontinuity (vertical asymptote). 2. $y = \frac{x^{2}-100}{x^{2}-12x + 20}$

Explanation:

Step1: Factor numerator and denominator

• Numerator: \(x^{2}-100=(x + 10)(x - 10)\) (using \(a^{2}-b^{2}=(a + b)(a - b)\) with \(a=x\) and \(b = 10\))
• Denominator: \(x^{2}-12x + 20=(x-2)(x - 10)\) (factoring quadratic \(ax^{2}+bx + c\), looking for two numbers that multiply to \(ac=20\) and add to \(b=-12\), which are \(-2\) and \(-10\))

Step2: Find discontinuities

Set denominator equal to zero: \((x - 2)(x - 10)=0\). Solutions are \(x=2\) and \(x = 10\)

Step3: Check nature of discontinuities

• For \(x = 10\):

Substitute \(x = 10\) into numerator: \((10 + 10)(10 - 10)=0\). Since numerator is also zero at \(x = 10\), there is a removable discontinuity (hole) at \(x = 10\)

• For \(x=2\):

Substitute \(x = 2\) into numerator: \((2 + 10)(2 - 10)=(12)(-8)=-96
eq0\). Since numerator is non - zero at \(x = 2\), there is an infinite discontinuity (vertical asymptote) at \(x = 2\)

Answer:

Discontinuities at \(x = 2\) (infinite discontinuity) and \(x = 10\) (removable discontinuity)