Question

1. reinforce

a. describe the transformations to the graph of ( y=\frac{1}{x} ) required to graph the function ( j(x)=\frac{-5}{x + 1}-4 ).
b. convert ( j(x)=\frac{-5}{x + 1}-4 ) to quotient form.
c. convert ( k(x)=\frac{7x + 3}{x - 2} ) to transformation form.

Explanation:

a.

Step1: Analyze the vertical stretch/compression and reflection

For the function \(y = \frac{1}{x}\) to \(y=\frac{- 5}{x + 1}-4\), the coefficient of \(\frac{1}{x+1}\) is \(-5\). The absolute value \(| - 5|=5\) means a vertical stretch by a factor of \(5\), and the negative sign means a reflection over the \(x\) - axis.

Step2: Analyze the horizontal translation

The function \(y=\frac{-5}{x + 1}-4\) is of the form \(y=\frac{a}{x - h}+k\) (where \(a=-5\), \(h=-1\), \(k = - 4\)). For the horizontal translation, if we compare \(y=\frac{1}{x}\) (where \(h = 0\)) to \(y=\frac{-5}{x+1}\) (where \(h=-1\)), using the rule \(y=\frac{1}{x}\to y=\frac{1}{x - h}\), we have a shift of \(1\) unit to the left.

Step3: Analyze the vertical translation

Using the rule \(y=\frac{a}{x - h}\to y=\frac{a}{x - h}+k\), when \(k=-4\), we have a shift of \(4\) units down.

Step1: Find a common denominator

We have \(j(x)=\frac{-5}{x + 1}-4\). The common denominator is \(x + 1\). So, \(j(x)=\frac{-5}{x+1}-\frac{4(x + 1)}{x + 1}\).

Step2: Expand and simplify the numerator

Expand \(4(x + 1)=4x+4\). Then \(j(x)=\frac{-5-(4x + 4)}{x + 1}=\frac{-5-4x - 4}{x + 1}=\frac{-4x-9}{x + 1}\)

Step1: Use polynomial long - division or rewrite the numerator

We rewrite \(7x+3\) as \(7(x - 2)+14 + 3\) (using the distributive property \(7(x - 2)=7x-14\)). So \(7x+3=7(x - 2)+17\)

Step2: Substitute into the function

Then \(k(x)=\frac{7(x - 2)+17}{x - 2}\). Using the rule \(\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}\) (where \(a = 7(x - 2)\), \(b = 17\), \(c=x - 2\)), we get \(k(x)=7+\frac{17}{x - 2}\)

Answer:

The graph of \(y = \frac{1}{x}\) is reflected over the \(x\) - axis, vertically stretched by a factor of \(5\), shifted \(1\) unit to the left and \(4\) units down.

b.