# Show that the minimum and maximum points of every curve in the family lieon the parabola y = 1 - x^2. Illustrate by graphing this parabola and severalmembers of the family.

The problem is incomplete since it does not provide the equation that represents the family of the curves whose extreme points lie on the parabola `y = 1 - x^2` .

Considering as example the family of curves `f_m(x) = ax^3 + bx^2 + cx + d` , you need to evaluate it's extreme points (maximum and minimum) and then you need to prove that these points lie on parabola `y = 1 - x^2` .

You need to find the extreme points of the family of curves given by the equation `f_m(x) = ax^3 + bx^2 + cx + d` , evaluating the solutions to the equation `(f_m(x))' = 0` , such that:

`(ax^3 + bx^2 + cx + d)' = 0 => 3ax^2 + 2bx + c = 0`

You need to evaluate the solutions to the equation `3ax^2 + 2bx + c = 0` , using quadratic formula, such that:

`x_(1,2) = (-2b +- sqrt(4b^2 - 12ac))/(6a)`

If `4b^2 - 12ac > 0` , the equation `3ax^2 + 2bx + c = 0` has two distinct solutions, hence, the function f_m(x) reaches its extremes at `x_(1,2) = (-2b +- sqrt(4b^2 - 12ac))/(6a)` .

If `4b^2 - 12ac = 0` , the equation `3ax^2 + 2bx + c = 0 ` has two equal solutions, hence, the function `f_m(x) ` reaches its extreme at `x = (-b)/(3a)` .

If `4b^2 - 12ac < 0` , the equation` 3ax^2 + 2bx + c = 0` has no real solutions, hence, the function `f_m(x)` has no extremes.

You need then to prove that either `x_(1,2) = (-2b +- sqrt(4b^2 - 12ac))/(6a)` , or `x = (-b)/(3a)` lie on parabola` y = 1 - x^2` .

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