# 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.

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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)` , orĀ `x = (-b)/(3a)` lie on parabola` y = 1 - x^2` .**