BQ #7: Unit V - Derivatives and the Area Problem

The formula for the difference quotient comes from a secant line intersecting twice, through a parabola. As it intersects through a parabola, two points are shared by the two lines. In order to work with tangent lines (lines that touch other lines upon only one point, and which we plot out with derivatives), we want the space between those two lines to be literally 0, meaning ideally, that we want to make them the same.

If we label them exactly as we were told to by a higher entity, then we label the further point as (x+h), and the closer point as just 'x', making the space in between them, 'h'. Their respective y-values, would be f (x+h) and f (x), which basically mean that they are plugged in and calculated into whatever equation we have, to get the y-value. Below is an example of what we have just done.

Somehow, we find it necessary to find the slope of the secant line; so, we utilize the slope formula of (y2 - y1) / (x2 - x1), which, if we plug in the values of our previous graph into, we get ( f (x+h) - f (x) ) / ( (x+h) - x) ) , which simplified, would be (f (x+h) - f (x) / h). And as we all know, that is our formula for the difference quotient, which is inextricably tied to derivatives, somehow.

To make the points (nearly) equal to each other, we can set the limit of h to 0, creating a situation where the points are not equal to each other, but extremely close to one another. After working with the difference quotient, using any sort of equation, we would arrive at an end where the variable 'h' probably exists. By making the limit of h equal to 0, we replace the secant line (two intersections), with the tangent line (one intersection), which are tied to derivatives by the fact that derivatives plot out the slope of every tangent line of any graph. Something like that.