Asked by pre calc
                f(x)=4x^2-8x+9
find the vertex the maximum the minimum the range and the intervals where it increases and decreases
            
        find the vertex the maximum the minimum the range and the intervals where it increases and decreases
Answers
                    Answered by
            drwls
            
    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
    
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
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