Question

c. determine the vertical asymptotes and removable discontinuities, if any, of $t(x)=\frac{2x^{2}-8x + 6}{x^{2}+x - 12}$. d. determine the asymptotes and removable discontinuities, if any, of $g(x)=\frac{5x}{x^{2}-5x}$. express the domain and range of $g(x)$ using interval notation, inequalities, and set notation.

Explanation:

Step1: Simplify the function \(g(x)\)

Factor the denominator: \(x^{2}-5x=x(x - 5)\). So \(g(x)=\frac{5x}{x(x - 5)}=\frac{5}{x - 5}\), \(x
eq0,x
eq5\)

Step2: Find removable discontinuities

Since \(x = 0\) makes the original function undefined but cancels out after simplification, there is a removable discontinuity at \(x = 0\)

Step3: Find vertical asymptotes

Set the simplified denominator \(x-5=0\), so \(x = 5\) is the vertical asymptote

Step4: Find the domain

• Inequalities: \(x

eq0\) and \(x
eq5\)

• Interval notation: \((-\infty,0)\cup(0,5)\cup(5,\infty)\)
• Set notation: \(\{x|x\in R,x

eq0,x
eq5\}\)

Step5: Find the range

Let \(y=\frac{5}{x - 5}\), then \(x=\frac{5}{y}+5\). Since \(y
eq0\)

• Inequalities: \(y

eq0\)

• Interval notation: \((-\infty,0)\cup(0,\infty)\)
• Set notation: \(\{y|y\in R,y

eq0\}\)

Answer:

• Removable discontinuity at \(x = 0\)
• Vertical asymptote at \(x = 5\)
• Domain:
• Inequalities: \(x

eq0,x
eq5\)

• Interval notation: \((-\infty,0)\cup(0,5)\cup(5,\infty)\)
• Set notation: \(\{x|x\in R,x

eq0,x
eq5\}\)

• Range:
• Inequalities: \(y

eq0\)

• Interval notation: \((-\infty,0)\cup(0,\infty)\)
• Set notation: \(\{y|y\in R,y

eq0\}\)