MA 109 Student Guide

1. To draw \(\frac{1}{2}p(x)\text{,}\) we see that we squishing the whole graph vertically in half (towards the \(x\)-axis). Since we are dividing all the \(y\)-values by 2, and \(0\) times anything is still \(0\text{,}\) all of the points on the \(x\)-axis (where \(y=0\)) stay where they are at. Next, we will move the corner points, then connect them together to get the whole graph. So, you would move the point \((-1,2)\) on the original graph down to \((-1,1)\text{.}\) Similarly, you would move the point \((2,-1)\) on the original graph up to \((2,-\frac{1}{2})\text{,}\) move the point \((3,-3)\) on the original graph up to \((3,-1.5)\text{,}\) and move the point \((6,3)\) on the original graph down to \((6,1.5)\text{.}\) From there, you can connect the lines to get the final graph, shown in blue below, where the gray dotted line is the orignal graph of \(p(x)\text{.}\)
   
2. To draw \(3p(x)\text{,}\) we see that we stretching the whole graph vertically by 3 (away from the \(x\)-axis). Since we are multiplying all the \(y\)-values by 3, and \(0\) times anything is still \(0\text{,}\) all of the points on the \(x\)-axis (where \(y=0\)) stay where they are at. Next, we will move the corner points, then connect them together to get the whole graph. So, you would move the point \((-1,2)\) on the original graph up to \((-1,6)\text{.}\) Similarly, you would move the point \((2,-1)\) on the original graph down to \((2,-3)\text{,}\) move the point \((3,-3)\) on the original graph down to \((3,-9)\text{,}\) and move the point \((6,3)\) on the original graph up to \((6,9)\text{.}\) From there, you can connect the lines to get the final graph, shown in blue below, where the gray dotted line is the orignal graph of \(p(x)\text{.}\)
   

We want to write a formula for \(g(x)\) in terms of \(f(x)\text{.}\) In other words, we start with \(f(x)\) and figure out how the graph changed to get to \(g(x)\text{,}\) then write how that changes the formula. Since the question asks for the formula "in terms of \(f(x)\)", that means we will write \(f(x)\) somewhere in our answer. Looking at the graph, we see that \(g(x)\) is the graph of \(f(x)\) stretched vertically (away from the \(x\)-axis) by \(2\text{.}\) Therefore, that means we are multiplying by \(2\) on the outside of the function \(f(x)\text{.}\) So, our final answer is \(g(x)=2f(x)\text{.}\)

Since we have an explicit formula for \(f(x)\) in this case, we can write an explicit formula for \(g(x)\text{,}\) too. Since we are squishing vertically by \(4\text{,}\) that means we are dividing by \(4\) on the outside of the function. So,

So our final answer is \(g(x)=\frac{3x^3-2x+1}{4}\text{.}\)

Comparing the two formulas, we see that in the formula for \(g(x)\text{,}\) all of the numbers are divided by \(5\text{.}\) That means that to get from \(f(x)\) to \(g(x)\text{,}\) we must have divided by \(5\) on the outside so that \(g(x)=\frac{f(x)}{5}\text{.}\) That means there was a vertical squish by \(5\text{.}\)