Question

Given a parabola `y = ax^2 + bx + c` find the slope of the parabola at the point (x, y) without using derivatives or any limits? is it possible to generalize for any function f(x)?

Answer 1

Yes, we can find it using infinitesimals...

Explanation:

We can find the slope using infinitesimals.

Let `epsilon` denote an infinitesimal.

Let `f(x) = ax^2+bx+c`

Then the slope at `x` is the standard part of:

`(f(x+epsilon) - f(x))/((x+epsilon) - x)`

`= ((a(x+epsilon)^2+b(x+epsilon)+c)-(ax^2+bx+c))/epsilon`

`= ((a(x^2+2epsilonx+epsilon^2)+b(x+epsilon)+c)-(ax^2+bx+c))/epsilon`

`= ((ax^2+(b+2epsilona)x+(c+bepsilon+aepsilon^2))-(ax^2+bx+c))/epsilon`

`= (2epsilonax+(bepsilon+aepsilon^2))/epsilon`

`= 2ax+b+aepsilon`

The standard part is obtained by discarding the infinitesimal part `aepsilon` to get `2ax+b`

The same method works for any differentiable function `f(x)`.

See https://socratic.org/s/aLLn6ic7 for more discussion.

Answer 2

See below.

Explanation:

Giving two points pertaining to the parabola, we have the slope between the points `p_1=(x_1,y_1), p_2 = (x_2,y_2)`

`(y_2-y_1)/(x_2-x_1) = (ax_2^2+bx_2+c -(a x_1^2+b x_1+c))/(x_2-x_1) = a(x_2+x_1)+b`

now making `x_1 = x_2 = x` we have

`y'(x) = 2ax+b`

Answer 3

Here's another method based on geometrical considerations...

Explanation:

Here's another method, without limits (explicit or not) or infinitesimals...

Given:

`f(x) = ax^2+bx+c`

Suppose `(x_1, y_1)` is some point on the parabola.

So:

`y_1 = ax_1^2+bx_1+c`

The equation of a non-vertical line through `(x_1, y_1)` can be written in the form:

`y-y_1 = m(x-x_1)`

i.e.:

`y = mx+(y_1-mx_1)`

This will be a tangent to the parabola if and only if the only intersection with the parabola is at `(x_1, y_1)`.

To find the points of intersection, we want to solve the system of equations:

`{ (y = ax^2+bx+c), (y = mx+(y_1-mx_1)) :}`

So:

`ax^2+bx+c = y = mx+ax_1^2+bx_1+c-mx_1`

That is:

`a(x^2-x_1^2)+b(x-x_1)-m(x-x_1) = 0`

i.e.:

`(x-x_1)(a(x+x_1)+b-m) = 0`

So if `x != x_1`, we have:

`a(x+x_1)+b-m = 0`

If the only solution is `x=x_1` then we can substitute `x_1` for `x` to find;

`2ax_1+b-m = 0`

So:

`m = 2ax_1+b`

We can potentially use a similar method with other functions, but the tangent to a parabola is guaranteed to touch the curve just once.