Question

1. use tables to investigate what happens to ( h(x) ) for values of ( x ) very near -3 and 2. complete the tables and describe what the values in each table tell you about the behavior of the graph.

a. for values near ( x=-3 ):

( x )	( h(x)=\frac{5x - 10}{x^{2}+x - 6} )
-3.1	-50
-3.01
-3.001
-3	undefined

b. for values near ( x = 2 ):

( x )	( h(x)=\frac{5x - 10}{x^{2}+x - 6} )
1.999
2	undefined
2.001

Explanation:

Step1: Calculate \(h(x)\) for \(x = - 3.01\)

Substitute \(x=-3.01\) into \(h(x)=\frac{5x - 10}{x^{2}+x - 6}\).

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Step2: Calculate \(h(x)\) for \(x=-3.001\)

Substitute \(x = - 3.001\) into \(h(x)=\frac{5x - 10}{x^{2}+x - 6}\).

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Step3: Calculate \(h(x)\) for \(x = 1.999\)

Substitute \(x = 1.999\) into \(h(x)=\frac{5x - 10}{x^{2}+x - 6}\).

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Step4: Calculate \(h(x)\) for \(x = 2.001\)

Substitute \(x = 2.001\) into \(h(x)=\frac{5x - 10}{x^{2}+x - 6}\).

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Step5: Analyze the behavior near \(x=-3\)

As \(x\) approaches \(-3\) from the left (\(x=-4,x = - 3.1,x=-3.01,x=-3.001\)), \(h(x)\) approaches \(-\infty\). The function \(h(x)\) has a vertical asymptote at \(x=-3\) since the denominator \(x^{2}+x - 6=(x + 3)(x - 2)\) and the numerator \(5x-10
eq0\) when \(x=-3\).

Step6: Analyze the behavior near \(x = 2\)

As \(x\) approaches \(2\) from the left (\(x = 1.999\)) and from the right (\(x=2.001\)), \(h(x)\) approaches \(1\). Since \(h(x)=\frac{5(x - 2)}{(x + 3)(x - 2)}=\frac{5}{x + 3}\) for \(x
eq2\), there is a removable discontinuity (a hole) at \(x = 2\)

Answer:

\(x\)	\(h(x)=\frac{5x - 10}{x^{2}+x - 6}\)	\(x\)	\(h(x)=\frac{5x - 10}{x^{2}+x - 6}\)
\(-3.1\)	\(-50\)	\(1.99\)	\(5.05\)
\(-3.01\)	\(-500\)	\(1.999\)	\(1\)
\(-3.001\)	\(-5000\)	\(2\)	Undefined
\(-3\)	Undefined	\(2.001\)	\(1\)
\(-2.999\)	\(5000\)	\(2.01\)	\(0.995\)
\(-2.99\)	\(500\)	\(2.1\)	\(0.909\)
\(-2.9\)	\(50\)	\(3\)	\(0.5\)

For \(x\) near \(-3\): The function \(h(x)\) has a vertical asymptote at \(x=-3\). As \(x\) approaches \(-3\) from the left, \(h(x)\to-\infty\). For \(x\) near \(2\): The function \(h(x)\) has a removable discontinuity (a hole) at \(x = 2\). As \(x\) approaches \(2\) from the left and the right, \(h(x)\to1\)