SOLUTION: f(x)=x^2-6x+12 g(x)=k In the equation above, f(x)>=g(x) for all real numbers x. If k is a constant, what is the maximum value of k?

Algebra ->  Expressions -> SOLUTION: f(x)=x^2-6x+12 g(x)=k In the equation above, f(x)>=g(x) for all real numbers x. If k is a constant, what is the maximum value of k?      Log On


   

Question 1043305: f(x)=x^2-6x+12
g(x)=k
In the equation above, f(x)>=g(x) for all real numbers x. If k is a constant, what is the maximum value of k?
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
f%3E=g
x%5E2-6x%2B12%3E=k
x%5E2-6x%2B12-k%3E=0

Maybe better, expect k to be the vertex, minimum value for f(x).

f%28x%29=x%5E2-6x%2B9-9%2B12 to complete-the-square;
%28x-3%29%5E2%2B3
Vertex is (3,3).
The line g(x)=k must be no higher than y=3. The MAXIMUM value for k is 3.