0=x^2-12x+39 find the y-intercept, and the domain and range

X^2-12x+39=0  

The y-intercept occurs when x=0, so in this case it is (0, 39), y=39

There are no restrictions upon x so the domain is all real numbers, or if you prefer...x=(-oo,+oo).

dy/dx=2x-12 and d2y/d2x=2

Since acceleration is always a positive 2, when velocity is equal to zero, y is at an absolute minimum...

dy/dx=0 only when 2x-12=0, 2x=12, x=6

y(6)=36-72+39=3

So the minimum y is 3 and y increases without bound as x approaches ±oo.

This means that the range is y=[3, +oo)