College Algebra

Example 11.57. Warm-up.

Suppose \(f(x)\) is given in the table below:

Now, let's define a new function \(g(x)=-f(x)\text{.}\) We want to construct the table for \(g(x)\text{:}\)

To fill in the top row, we need to find \(g(-1)\text{.}\) Based on our definition of \(g(x)\text{,}\) we have that \(g(-1)=-f(-1)\text{.}\) From the table for \(f(x)\text{,}\) we know that \(f(-1)=2\text{.}\) Therefore,

We can fill out the rest of the table following the same logic:

Now, let's define a new function \(h(x)=f(-x)\text{.}\) We want to construct the table for \(h(x)\text{.}\) Let's follow the same logic we had for \(g(x)\) above: to fill in the top row, we need to find \(h(-1)\text{.}\) Based on our definition of \(h(x)\text{,}\) we have that \(h(-1)=f(-(-1))=f(1)\text{.}\) However, at this point we are stuck, because \(x=1\) is not in our table for \(f(x)\text{.}\) So, we cannot use the same input values for \(h(x)\text{.}\) Instead, we need to reverse engineer what inputs we can use for \(h(x)\) that will match up to the inputs we have for \(f(x)\text{.}\) For example, we know what \(f(-1)\) is, and \(h(1)=f(-(1))=f(-1)\text{,}\) that means we can use \(1\) as one of our inputs for \(h(x)\) and we have that \(h(1)=2\text{.}\) We can follow the same logic to get the rest of the table:

Example 11.65.

Below is the graph of \(h(x)\text{:}\)

We want to figure out which of the following is the graph of \(-h(x)\text{.}\) In each, the gray dotted graph is the orignal graph of \(h(x)\) so you can more easily see how it changed.

Since we are looking for \(-h(x)\text{,}\) we are looking for the graph to flip vertically (over the \(x\)-axis), which is graph C. Graph A flipped horizontally over the \(y\)-axis, which would be \(h(-x)\text{.}\) Graph B shifted down by 1, which we learned in Section 11.1 would be \(h(x)-1\text{.}\) Graph D is flipped upside down, but it's too high on the graph, since we want to flip over the \(x\)-axis. We'll learn about how to combine a flip and a shift in Section 11.6

Example 11.67.

Suppose \(f(x)=x^7-2x^4+5x^3-2x+6\) and \(g(x)\) is the same as \(f(x)\) but flipped vertically. Let's write a formula for \(g(x)\text{.}\) Since we are flipping vertically, that means we are multiplying by \(-1\) on the outside of the function.

What if we wanted to flip it horizontally instead? In that case, we would be multiplying by \(-1\) on the inside of the function.

Notice how we simplified in the equation above. Any negatives inside of an even power cancel out and don't do anything. However, the negatives in the odd powers do not cancel, so they flip signs.