4.4 Five Graphs: Slope
Let's revisit our five graphs but this time wearing glasses that see slope!
Squaring
We've already spent a little time looking at the squaring function,
\[
\begin{align}
z &\depends x
\\
z &= x^2,
\end{align}
\]
and its slope
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= 2 x.
\end{align}
\]
Let's draw a transition diagram for the slope. The parabola has a level point at \( x = 0 \). In particular, this is a low point for the graph. The low point appears in the transition diagram: the graph transitions from indirect variance to direct variance at \( x = 0 \).
Cubing
Recall the graph of the cubing function,
\[
\begin{align}
z &\depends x
\\
z &= x^3.
\end{align}
\]
The cubing function has slope
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= 3 x^2.
\end{align}
\]
We find a level point at \( x = 0 \), and the slope is positive at all other points. The cubing function gives us an interesting example of a level point! We find larger heights when moving to right, and we find smaller heights by moving to the left. So this level point at is neither a low point nor a high point for the graph.
Reciprocal
The reciprocal function,
\[
\begin{align}
z &\depends x
\\
z &= \frac{1}{x},
\end{align}
\]
is defined at all non-zero inputs, \( x \neq 0 \). Here's the slope.
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= \frac{-1}{x^2}
\end{align}
\]
The slope is negative wherever it is defined. In other words, the graph has indirect variance.
Square Root
Next let's look at the square root function,
\[
\begin{align}
z &\depends x
\\
z &= \sqrt[2]{x}.
\end{align}
\]
The square root has slope
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= \frac{1}{2 \mult \sqrt[2]{x}}.
\end{align}
\]
The graph has direct variance at all allowed inputs. We must disallow \( x = 0 \), however, for the slope to be defined.
Cube Root
Last let's check in with the cube root function,
\[
\begin{align}
z &\depends x
\\
z &= \sqrt[3]{x}.
\end{align}
\]
We've calculated the slope to be
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= \frac{1}{3 \mult \bigl(\sqrt[3]{x}\bigr)^2}.
\end{align}
\]
Once again, we must disallow the input \( x = 0 \) for the slope to be defined. The graph has direct variance at all allowed points.
