Let D(n) denote change in some integer n; let A(x) and A(y) denote either the corresponding x or y co-ordinate at some point A on the euclidean plane;
Question: Considering two points P and Q upon the curve y= x^2-2x-3 of the euclidean plane what is the slope of the secant line PQ at the fixed point P as Q approaches P?
slope m of secant line PQ = D(y)/D(x)
so, if D(y)= Q(y) - P(y) then Q(y)= P(y) + D(y)
and if D(x)=Q(x) - P(x) then Q(x)=P(x) + D(x)
i.e. D(y) = [x^2 + 2xD(x) + D(x)^2 - 2x - 2D(x) - 3] - [x^2 - 2x - 3]
and furthermore D(y)/D(x)= 2x + D(x) - 2
am uncertain of final solution and would like to know if this answer implies the limit of the slope of the secant line as D(x) approaches 0 is 2x - 2
The problem asked what happens "as Q approaches P", which means, yes, D(x) goes to 0 and the the slope of the secant line goes to 2x- 2. That, by the way, implies that the equation of the tangent line, at is .
Originally Posted by Foxlion
I presume that this is the very beginning of a Calculus or Analysis course and you will so be introduced to the "derivative" as "slope of the tangent line".
you presume correct, thank you.
Originally Posted by HallsofIvy