Question

1. how do you determine where a rational function has a:

a. discontinuity
b. vertical asymptote
c. removable discontinuity
d. x-intercept
e. y-intercept

Explanation:

• a. Discontinuity: A rational function \(f(x)=\frac{p(x)}{q(x)}\) has a discontinuity at values of \(x\) where \(q(x) = 0\).
• b. Vertical Asymptote: After simplifying the rational function \(f(x)=\frac{p(x)}{q(x)}\) (by canceling common factors in \(p(x)\) and \(q(x)\)), vertical asymptotes occur at the values of \(x\) that make the simplified denominator equal to \(0\).
• c. Removable Discontinuity: If the rational function \(f(x)=\frac{p(x)}{q(x)}\) can be simplified by canceling a common factor \((x - a)\) (where \(a\) is a real number) from both \(p(x)\) and \(q(x)\), then there is a removable discontinuity at \(x=a\).
• d. \(x\)-Intercept: Set the numerator \(p(x)=0\) (while ensuring that the denominator \(q(x)

eq0\) for those \(x\) - values) and solve for \(x\). The solutions of \(p(x) = 0\) (with \(q(x)
eq0\)) are the \(x\) - intercepts.

• e. \(y\)-Intercept: Substitute \(x = 0\) into the rational function \(y=f(0)=\frac{p(0)}{q(0)}\), provided \(q(0)

eq0\).

Answer:

• a. At \(x\) - values where the denominator is \(0\).
• b. At \(x\) - values that make the simplified denominator \(0\).
• c. When a common factor can be canceled from numerator and denominator.
• d. By solving \(p(x)=0\) (\(q(x)

eq0\)).

• e. By evaluating \(f(0)\) (\(q(0)

eq0\)).