Question

c. determine the vertical asymptotes and removable discontinuities, if any, of ( t(x)=\frac{2 x^{2}-8 x + 6}{x^{2}+x - 12} ).

Explanation:

Step1: Factor numerator and denominator

• Numerator: \(2x^{2}-8x + 6=2(x^{2}-4x + 3)=2(x - 1)(x - 3)\)
• Denominator: \(x^{2}+x - 12=(x + 4)(x - 3)\)

Step2: Simplify the function

\(t(x)=\frac{2(x - 1)(x - 3)}{(x + 4)(x - 3)}=\frac{2(x - 1)}{x + 4}\), \(x
eq3\)

Step3: Find vertical asymptote

Set the simplified denominator \(x + 4=0\), solve for \(x\). We get \(x=-4\)

Step4: Find removable discontinuity

The factor \((x - 3)\) cancels out. So there is a removable discontinuity at \(x = 3\)

Answer:

Vertical asymptote: \(x = - 4\); Removable discontinuity at \(x = 3\)