Try completing the square first.
4x^2 -8x +9 = 4(x^2 -2x +1) +9 -4
= 4(x-1)^2 + 5
The vertex is where it has its lowest value, which is obviously at x=1, f(x) = 5
The function increases for x<1 and x>1
There is no maximum.
The range is 5 to infinity
f(x)=4x^2-8x+9
find the vertex the maximum the minimum the range and the intervals where it increases and decreases
1 answer