Applied Algebra: Modeling and functions

1. The table shows that the \(y\)-values for \(g(x)\) are each three units greater than the corresponding \(y\)-values for the absolute value function. The graph of \(g(x) = \abs{x} + 3\) is a translation of the basic graph of \(y = \abs{x}\text{,}\) shifted upward three units, as shown below.

   

   \(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
   \(y=\abs{x}\)	\(2\)	\(1\)	\(0\)	\(1\)	\(2\)
   \(g(x)=\abs{x}+3\)	\(5\)	\(4\)	\(3\)	\(4\)	\(5\)

2. The table shows that the \(y\)-values for \(h(x)\) are each two units smaller than the corresponding \(y\)-values for \(y = \dfrac{1}{x}\text{.}\) The graph of \(h(x) = \dfrac{1}{x} - 2\) is a translation of the basic graph of \(y = \dfrac{1}{x}\text{,}\) shifted downward two units, as shown below.

   

   \(x\)	\(-2\)	\(-1\)	\(\dfrac{1}{2}\)	\(1\)	\(2\)
   \(y=\dfrac{1}{x}\)	\(\dfrac{-1}{2}\)	\(-1\)	\(2\)	\(1\)	\(\dfrac{1}{2}\)
   \(h(x)=\dfrac{1}{x}-2\)	\(\dfrac{-5}{2}\)	\(-3\)	\(0\)	\(-1\)	\(\dfrac{-3}{2}\)

The graph of \(y = f (h) + 300\) is a vertical translation of the graph of \(f\text{,}\)

as shown at right. At each hour of the day, or for each value of \(h\text{,}\) the \(y\)-coordinate is 300 greater than on the graph of \(f\text{.}\) So at each hour, the community is drawing 300 megawatts more power than in 2002.

1. Consider the table of values for the function.

   \(x\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)
   \(y=\sqrt{x}\)	undefined	\(0\)	\(1\)	\(1.414\)	\(1.732\)
   \(y=\sqrt{x+1}\)	\(0\)	\(1\)	\(1.414\)	\(1.732\)	\(2\)

   The table shows that each \(y\)-value for \(g(x)\) occurs one unit to the left of the same \(y\)-value for the graph of \(y=\sqrt{x}\text{.}\) Consequently, each point on the graph of \(y = g(x)\) is shifted one unit to the left of \(y =\sqrt{x}\text{,}\) as shown at right.

   

2. Consider the table of values for the function.

   \(x\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
   \(y=\dfrac{1}{x}\)	\(1\)	undefined	\(1\)	\(\dfrac{1}{4}\)	\(\dfrac{1}{9}\)	\(\dfrac{1}{16}\)
   \(y=\dfrac{1}{(x-3)^2}\)	\(\dfrac{1}{16}\)	\(\dfrac{1}{9}\)	\(\dfrac{1}{4}\)	\(1\)	undefined	\(1\)

   The table shows that each \(y\)-value for \(h(x)\) occurs three units to the right of the same \(y\)-value for the graph of \(y =\dfrac{1}{x^2}\text{.}\) Consequently, each point on the graph of \(y = h(x)\) is shifted three units to the right of \(y =\dfrac{1}{x^2}\text{,}\) as shown at right.

   

1. The graph of \(g\) is translated \(10\) units to the right of \(f\text{,}\) so \(g(p) = f (p - 10)\text{.}\)
2. Pressure readings above \(40\) are a strong indication of glaucoma. Readings between \(10\) and \(40\) cannot conclusively distinguish healthy eyes from those with glaucoma.

We identify the basic graph from the structure of the formula for \(f (x)\text{.}\) In this case, the basic graph is \(y = x^3\text{,}\) so we begin by locating a few points on that graph, say, \((0,0),~(1,1),\) and \((-1,-1).\)

We'll perform the translations separately, following the order of operations. First, we sketch a graph of \(y = (x + 4)^3\) by shifting each point on the basic graph four units to the left. We then move each point up two units to obtain the graph of \(f(x) = (x + 4)^3 + 2\text{.}\) All three graphs are shown below.

1. The graph of \(g(x) = 3\sqrt[3]{x}\) is a vertical expansion of the basic graph \(y = \sqrt[3]{x}\) by a factor of \(3\text{,}\) as shown below . Each point on the basic graph has its \(y\)-coordinate tripled.

   

2. The graph of \(h(x) = \dfrac{-1}{2}\abs{x}\) is a vertical compression of the basic graph \(y = \abs{x}\) by a factor of \(\dfrac{1}{2}\text{,}\) combined with a reflection about the \(x\)-axis. You may find it helpful to graph the function in two steps, as shown below.

   

To sketch a graph of \(g\text{,}\) we stretch the graph of \(f\) vertically by a factor of \(2\text{,}\) as shown below. At each time \(t\text{,}\) the person’s blood alcohol level is twice the value given by \(f\text{.}\) The function \(g\) could represent a person's blood alcohol level \(t\) hours after drinking two martinis.