Question

1 fill in the blanks with the correct values to describe features of the function $f(x)=\frac{x^{2}-5x + 6}{x - 2}$. for each answer, show your work or explain your reasoning.
a the function has a removable discontinuity at $x = a$.
b the function has a $y$-intercept at $(0,b)$.
c the function has an $x$-intercept at $(c,0)$.
d the function has a vertical asymptote at $x = d$.
e the function has a horizontal asymptote at $y = e$.

Explanation:

Step1: Simplify the function

Factor the numerator \(x^{2}-5x + 6=(x - 2)(x - 3)\). So \(f(x)=\frac{(x - 2)(x - 3)}{x - 2}\). For \(x
eq2\), \(f(x)=x - 3\). A removable discontinuity occurs when a factor cancels out. Here, the factor \((x - 2)\) cancels, so \(A = 2\).

Step2: Find the \(y\)-intercept

The \(y\)-intercept is found by setting \(x = 0\). Since \(f(x)=x - 3\) (for \(x
eq2\)), \(f(0)=0-3=-3\). So \(B=-3\).

Step3: Find the \(x\)-intercept

The \(x\)-intercept is found by setting \(y = 0\). Set \(x - 3=0\), then \(x = 3\). So \(C = 3\).

Step4: Check for vertical asymptote

Since the simplified function \(y=x - 3\) (a linear function) has no vertical asymptote (the original discontinuity at \(x = 2\) is removable), \(D\) does not exist (or we can say there is no vertical asymptote).

Step5: Check for horizontal asymptote

Since \(y=x - 3\) (a linear function \(y=mx + b\) with \(m = 1
eq0\)), there is no horizontal asymptote.

Answer:

a. \(A = 2\)
b. \(B=-3\)
c. \(C = 3\)
d. No vertical asymptote (or \(D\) is not applicable)
e. No horizontal asymptote (or \(E\) is not applicable)