# Linear, Quadratic, and Cubic Functions

These are names for functions of first, second and third order polynomial functions, respectively. What this means is that the highest order of x (the variable) in the function is 1, 2 or 3.

The generalized form for a linear function (1 is highest power):
f(x) = ax+b, where a and b are constants, and a is not equal to 0

The generalized form for a quadratic function (2 is highest power):
f(x) = ax2+bx+c, where a, b and c are constants, and a is not equal to 0

The generalized form for a cubic function (3 is highest power):
f(x) = ax3+bx2+cx+d, where a, b, c and d are constants, and a is not equal to 0

The roots of a function are defined as the points where the function f(x)=0. For linear and quadratic functions, this is fairly straight-forward, but the formula for a cubic is quite complicated and higher powers get even more involved. I will go over the derivation of the first two now, But the rest plus a history of the discoveries can be found here.

A linear equation is very simple to solve for f(x)=0:

```0   = ax+b
-ax = b
```
` x  = b/(-a) = -b/a`
`,  a not equal to 0`

The equation for the root of a quadratic is only slightly more complex. The idea is to isolate x by putting the left side into the form (x+q)2 and then taking the square root. We do this by some nifty algebra:
 ax2 + bx + c = 0 x2 + (b/a)x + c/a = 0 [x2 + (b/a)x + b2/4a2] - b2/4a2 + c/a = 0 [x + (b/2a)]2 - ( b2/4a2 - c/a) = 0 [x + (b/2a)]2 = b2/4a2 - c/a x+(b/2a) = sqrt(b2/4a2-4ac/4a2) x = -b/2a sqrt(b2-4ac)/2a x+(b/2a) = sqrt(b2/4a2-4ac/4a2) x = (-b sqrt(b2-4ac))/2a Try to get (x+g)2 = x2 + (b/a)x + ?? (x+(1/2)(b/a))2 = x2 + 2(1/2)(b/a)x + (1/4)(b2/a2) (x+b/2a)2 = x2 + (b/a)x + b2/4a2

-David