Question

standard: f - ifc.07d graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. (d) (+) graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior.
1 fill in the blanks with the correct values to describe features of the function ( f(x)=\frac{x^{2}-5 x + 6}{x - 2} ). for each answer, show your work or explain your reasoning.
a the function has a removable discontinuity at ( x = a ).

Explanation:

Step1: Factor the numerator

Factor \(x^{2}-5x + 6\). We know that \(x^{2}-5x + 6=(x - 2)(x-3)\) (using the formula \(ax^{2}+bx + c=a(x - x_1)(x - x_2)\), where \(x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or by simple factoring: looking for two numbers that multiply to \(6\) and add up to \(-5\), which are \(-2\) and \(-3\)). So \(f(x)=\frac{(x - 2)(x - 3)}{x - 2}\)

Step2: Analyze the removable discontinuity

A removable discontinuity (hole) occurs when a factor in the numerator and a factor in the denominator cancel out. The function \(y = f(x)\) is undefined when the denominator is zero. For \(y=\frac{(x - 2)(x - 3)}{x - 2}\), when \(x-2 = 0\) (i.e., \(x = 2\)), the original function \(f(x)=\frac{x^{2}-5x + 6}{x - 2}\) is undefined. But since we can cancel the \((x - 2)\) terms (for \(x
eq2\)), \(x = 2\) is a removable discontinuity.

Answer:

\(2\)