# AoMath: Slope Of Tangent Lines To Curves

Slope of the tangent line. Pretty odd thing to study you might say. Let's start by drawing a continous and curvey function in the Cartesian plane from (1, 1) to (4, 2). Look to see if there are any "hills and valleys." Notice that the slope of the tangent line at the very bottom and very top is ZERO! So:

1. We could find a function that tells us the slope at every x-value.
2. We could set it equal to zero to see which x-values have a slope of zero.
3. We could then use those x-values to help us find the tallest hill or the lowest valley.
But we have a ways to go before we can do all that.
So how do we find the slope of a tangent line, say at x = 1/2, of $$f(x) = 1 - x^2$$? First, we need a point on the graph at which we want to know the slope.

1. We find the point (c, f(c)) or (1/2, 3/4).
2. We draw the secant slope between the point above and any OTHER point on the function, in our case $$(x_0, f(x_0))$$.
3. We draw more and more secant lines but move that second point closer to (1/2, 3/4) until we are VERY CLOSE but NOT EQUAL TO the original point. Sounds like a limit! $$f'(c) = \lim_{x_0 \rightarrow c} \frac{f(x_0) - f(c)}{x_0 - c} = ??$$ $$\lim_{x_0 \rightarrow \frac{1}{2}} \frac{f(x_0) - f(\frac{1}{2})}{x_0 - \frac{1}{2}}$$ $$= \lim_{x_0 \rightarrow \frac{1}{2}} \frac{(1 - (x_0)^2) - (1 - (\frac{1}{2})^2)}{x_0 - \frac{1}{2}}$$ $$= \lim_{x_0 \rightarrow \frac{1}{2}} \frac{(\frac{1}{4} - (x_0)^2)}{x_0 - \frac{1}{2}}$$ $$= \lim_{x_0 \rightarrow \frac{1}{2}} \frac{(1 - 4(x_0)^2)}{4x_0 - 2}$$ $$= \lim_{x_0 \rightarrow \frac{1}{2}} \frac{(1 - 2(x_0))(1 + 2(x_0))}{2(2x_0 - 1)}$$ $$= \lim_{x_0 \rightarrow \frac{1}{2}} \frac{-(1 + 2(x_0))}{2} = \boxed{-1}$$