Easier way to find the minimum value of a quadratic equations

GRE Quant Problem Solving
1

Warning !! - This technique requires an understanding of derivatives.

In problems where you need to find the minimum value such as module 6 - Graphing Quadratic equation, there is an easier way using derivatives.

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For example if you want to find the minimum of this quadratic, all you have to do is find the first derivative and equate it to 0.

First derivative: Y’ = 2x -2 = 0
This gives us x = 1. Substituting x = 1 in the original equation,
y = (1)2 -2(1) – 8
y = -9

Therefore, the minimum point of the parabola is (1,-9).

NOTE : The sign of x^2 has to be + for this to work.

Hope this helps
(:

3

I don’t think this is any faster than directly computing f\left( -\frac{b}{2a}\right) or completing the square.

This part isn’t really necessary. If the coefficient of x^2 is negative, the method still works; it simply returns the maximum instead of the minimum. Essentially, the point of interest is the vertex, and computing the derivative will return the x-coordinate of the vertex in all cases.

1 Like
4

They are actually equivalent. Take a general quadratic y = ax^2+bx+c where a is nonzero

Then, y’ = 2ax+b. Set that to 0: 2ax + b= 0 → 2ax = -b → x = -b/2a.

You are essentially calculating -b/2a and then gonna plug it in