Add content for functions/slopes-lines-curves

Add actual content to the skeleton of the
`functions/slopes-lines-curves` section.

Signed-off-by: Eggert Karl Hafsteinsson <[email protected]>
Signed-off-by: Teodor Dutu <[email protected]>
Signed-off-by: Razvan Deaconescu <[email protected]>

chapters/functions/slopes-lines-curves/reading/read.md

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+# Slopes of Lines and Curves
+
+## The Slope of a Line
+
+Linear functions produce straight-line graphs.
+In general, a straight line follows the following equation:
+
+$$y = a + bx$$
+
+where $a$ and $b$ are fixed numbers.
+
+The line on the graph is the set of points:
+
+$$\left[ (x,y): x,y \in \mathbb{R}, y = a+bx \right]$$
+
+![Fig. 22](../media/13_1_The_slope_of_a_line.png)
+
+### Details
+
+The slope of a straight line represents the change in the $y$ coordinate corresponding to a unit change in the $x$ coordinate.
+
+## Segment Slopes
+
+Let's assume we have a more general function $y = f(x)$.
+To find the slope of a line segment, consider two $x$ -coordinates ($x_0$ and $x_1$), and look at the slope between $(x_0, f(x_0))$ and $(x_1, f(x_1))$.
+
+![Fig. 23](../media/13_2_segment_slopes.png)
+
+### Details
+
+Consider two points, $(x_0,y_0)$ and $(x_1,y_1)$.
+The slope of the straight line that goes through these points is
+
+$$\displaystyle\frac {y_1 - y_0} {x_1 - x_0}$$
+
+Thus, the slope of a line segment passing through the points $(x_0,f(x_0))$ and $(x_1,f(x_1))$, for some function, $f$, is
+
+$$\displaystyle\frac {f(x_1) - f(x_0)} {x_1 - x_0}$$
+
+If we let $x_1 = x_0 + h$ then the slope of the segment is
+
+$$\displaystyle\frac {f(x_0+h) - f(x_0)} {h}$$
+
+## The Slope of $y=x^2$
+
+Consider the task of computing the slope of the function $y=x^2$ at a given point.
+
+![Fig. 24](../media/13_3_The_slope_of.png)
+
+### Examples
+
+Consider the function $y = f(x) = x^2$.
+In order to find the slope at a given point $(x_0 )$, we look at
+
+$$y = \displaystyle\frac{f (x_0 +h) - f(x_0)} {h}$$
+
+for small values of $h$.
+
+For this particular function, $f (x) = x^2$, and hence
+
+$$f (x_0 +h) = (x_0 +h) ^2 = x^2 + 2hx_0 + h^2$$
+
+The slope of a line segment is therefore given by
+
+$$\displaystyle\frac{f (x_0 +h) - f(x_0)} {h}= \displaystyle\frac{2hx_0 + h^2} {h} = 2x_0 + h$$
+
+As we make $h$ steadily smaller, the segment slope, $2x_0 + h$, tends towards $2x_0$.
+It follows that the slope, $y'$, of the curve **at a general point** $x$ is given by $y' = 2x$.
+
+## The Tangent to a Curve
+
+A **tangent** to a curve is a line that intersects the curve at exactly one point.
+The slope of a tangent for the function $y=f(x)$ at the point $(x_0,f(x_0))$ is
+
+$$\lim_{h\to0}\displaystyle\frac{f(x_0+h)-f(x_0)}{h}$$
+
+![Fig. 25](../media/13_4_The_tangent_to_a_curve.png)
+
+### Details
+
+To find the slope of the tangent to a curve at a point, we look at the slope of a line segment between the points $(x_0,f(x_0))$ and $(x_0+h,f(x_0+h))$, which is
+
+$$\displaystyle\frac{f(x_0+h)-f(x_0)}{h}$$
+
+and then we take $h$ to be closer and closer to $0$.
+Thus the slope is
+
+$$\lim_{h\to0}\displaystyle\frac{f(x_0+h)-f(x_0)}{h}$$
+
+when this limit exists.
+
+### Examples
+
+:::info Example
+
+We wish to find the tangent line for the function $f(x)=x^2$ at the point $(1,1)$.
+First we need to find the slope of this tangent, it is given as
+
+$$\lim_{h\to0}\displaystyle\frac{(1+h)^2-1^2}{h}=\lim_{h\to0}\displaystyle\frac{2h+h^2}{h}=\lim_{h\to0}(2+h)=2$$
+
+Then, since we know the tangent goes through the point $(1,1)$ the line is $y=2x-1$.
+
+:::
+
+## The Slope of a General Curve
+
+![Fig. 26](../media/13_5_The_slope_of_a_general_curve.png)
+
+### Details
+
+Imagine a nonlinear function whose graph is a curve described by the equation $y = f(x)$.
+Here we want to find the slope of a line tangent to the curve at a specific point $(x_0)$.
+The slope of the line segment is given by the equation $\displaystyle\frac{f (x_0 +h) - f(x_0)} {h}$.
+Reducing $h$ towards zero, gives the slope of this curve if it exists.