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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