Week 13 in Precalc 11

This week in math we learned how to graph linear reciprocal functions. The parent function would be in the format: f(x)=mx+b, but the reciprocal of that would look like f$(x)^{-1}$=$\frac{1}{mx+b}$. To graph this, the first step would be to graph the original parent function. So, as we learned before, use the slope and y-intercept to graph that linear line.

Next, We need to find the invariant points. The invariant points are the points on the graph which are ‘nice points’ so they aren’t fractions. To find these points, you look at the y-axis where y=1 and y=-1. Go across from the y-axis at 1 and -1 to where you bump into the line again. This would allow you to find the x value of the invariant points and therefore the location.

Once you have the invariant points, you need to find the asymptotes. The horizontal asymptote is the line that is directly between the two invariant points horizontally, show this by drawing a dotted line on it. In Precalculus 11, we won’t need to worry about this too much as it will most likely always equal 0. The vertical asymptote is the line that is directly between the two invariant points vertically, you can also find this asymptote by finding the x-intercept of the original parent function, as vertical asymptote = x-intercept of f(x). Also, show this by drawing another dotted line where this asymptote is.

Now that you have the parent function graphed, the invariant points, and the asymptotes, you can now graph the reciprocal function. This function will be a hyperbola, which means the line will come in two different parts. It’s hard to graph these lines accurately by hand, as they are two curved lines that keep getting closer and closer to zero, but never reach it on the asymptotes. Simply try to be as accurate as you can. I’ve included an example below to help visualize these steps.