The following equalities provide context to the definition of the tangent function. Now, what if $y$ is a function symbolized by $y(x)$ such that $y(x)=x$? In this case, $x$ is the independent variable and $y$ is the dependent variable. Then $ \tan\phi = \frac{x}{x}=1$. Interesting… 1 is also the slope of the linear function $ y(x)$. But of course it is the slope–after all, $ \tan\phi$ is defined as $ \frac{y}{x}$. Suddenly, this simple function is very interesting. In certain cases, $ \tan\phi = \frac{\Delta y}{\Delta x}$, which is a slope of a linear curve. Another example is if $ y = -x$. Then $ \tan\phi = -1$. This is the slope of the line $ y = -x$ which happens to intersect the origin. If the function does not intersect the origin, such as $ y(x) = x + 2$, then $ \tan\phi = \frac{x+2}{x} = 1 + \frac{2}{x}$. The slope of 1 is still on the right side, but there is a rather strange $ \frac{2}{x}$ part on the right side now. So $ y$-intercepts kind of ruin the nice relation between slopes (or…tangent lines) and the tangent function, unless there’s something I’m missing.