Question

1. how does the graph of the parent function compare to the graph of the bricklayers function, ( y=\frac{100}{x} )? how do the domains of the mathematical function and the function representing the problem situation compare?
2. reinforce use your knowledge of transformations of functions to graph the function ( h(x)=-\frac{5}{x} ). describe the strategy you used.

Explanation:

Step1: Identify the parent function

The parent function of \(y =-\frac{5}{x}\) is \(y=\frac{1}{x}\).

Step2: Analyze the transformation

For the function \(h(x)=-\frac{5}{x}\), compared to \(y = \frac{1}{x}\), there is a vertical stretch by a factor of \(5\) (because of the coefficient \(5\)) and a reflection over the \(x\) - axis (because of the negative sign).

Step3: Graphing strategy

1. Plot key points of the parent function:

• For \(y=\frac{1}{x}\), when \(x = 1\), \(y = 1\); when \(x=-1\), \(y=-1\).

1. Apply the transformation:

• For the vertical stretch: If \((x,y)\) is a point on \(y=\frac{1}{x}\), then for \(y = 5\times\frac{1}{x}\), the point becomes \((x,5y)\). So \((1,1)\) becomes \((1,5)\) and \((-1,-1)\) becomes \((-1, - 5)\).
• For the reflection over the \(x\) - axis: If \((x,y)\) is a point on \(y = 5\times\frac{1}{x}\), then for \(y=-5\times\frac{1}{x}\), the point becomes \((x,-y)\). So \((1,5)\) becomes \((1,-5)\) and \((-1,-5)\) becomes \((-1,5)\).

1. Draw the hyperbola:

• The function \(y =-\frac{5}{x}\) is a hyperbola. As \(x\to0^{+}\), \(y\to-\infty\); as \(x\to0^{-}\), \(y\to\infty\); as \(x\to\infty\), \(y\to0^{-}\); as \(x\to-\infty\), \(y\to0^{+}\).

Answer:

To graph \(h(x)=-\frac{5}{x}\), start with the parent function \(y = \frac{1}{x}\). Apply a vertical stretch by a factor of \(5\) (multiply \(y\) - values of points on \(y=\frac{1}{x}\) by \(5\)) and then a reflection over the \(x\) - axis (change the sign of \(y\) - values of points on \(y = 5\times\frac{1}{x}\)). Plot key points like \((1,-5)\), \((-1,5)\) and draw the hyperbola approaching the \(x\) and \(y\) axes (asymptotes \(x = 0\) and \(y=0\)).