Question

1. reinforce

a. convert ( h ( x ) = \frac { - 2 } { x + 3 } + 1 ) to quotient form. use a graph to confirm that the new form is equivalent.
b. convert ( v ( x ) = \frac { - 6 x + 11 } { x - 1 } ) to transformation form. use a graph to confirm that the new form is equivalent.

Explanation:

Step1: Combine the terms for \(h(x)\)

To convert \(h(x)=\frac{-2}{x + 3}+1\) to quotient form, we need to get a common denominator.

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To confirm graphically, we can use a graphing utility. The vertical asymptote of \(y=\frac{-2}{x + 3}+1\) is \(x=-3\) (from the denominator \(x + 3=0\)) and the horizontal asymptote is \(y = 1\) (since the degree of the numerator is equal to the degree of the denominator for the original rational - function part \(\frac{-2}{x+3}\), and then we have a vertical shift of \(1\)). For \(y=\frac{x + 1}{x+3}=\frac{x+3-2}{x + 3}=1-\frac{2}{x+3}\), it also has a vertical asymptote \(x=-3\) and a horizontal asymptote \(y = 1\).

Step2: Use polynomial long - division for \(v(x)\)

For \(v(x)=\frac{-6x + 11}{x-1}\), using polynomial long - division:

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So \(v(x)=\frac{-6x + 11}{x-1}=-6+\frac{5}{x - 1}\)
To confirm graphically, for \(y=\frac{-6x + 11}{x-1}\), the vertical asymptote is \(x = 1\) (from \(x-1=0\)) and the horizontal asymptote is \(y=-6\) (since the degree of the numerator and denominator are equal, and the ratio of the leading coefficients is \(\frac{-6}{1}=-6\)). For \(y=-6+\frac{5}{x - 1}\), it has a vertical asymptote \(x = 1\) and a horizontal asymptote \(y=-6\)

Answer:

a. \(h(x)=\frac{x + 1}{x+3}\)
b. \(v(x)=-6+\frac{5}{x - 1}\)