if f(x)=x^2+5x+6
then
-2f(x/11) = -2( (x/11)^2 + 5(x/11) + 6)
you want the zeros of that ...
-2[x^2/121 + 5x/11 + 6] = 0
x^2/121 + 5x/11 + 6 = 0 , I think you had it that far
times 121
x^2 + 55x + 726 = 0
(x+22)(x+33) = 0
x = -22 or x = -33
The function f(x)=x^2+5x+6 has zeros at -2 and -3. What are the zeros of the function y=-2f(x/11) ?
Do you just plug x/11 into each x value in f(x)? Like f(x)= (x/11)^2+5(x/11)+6 (I exclueded the -2 because if I divide 0 by -2, it's still zero)?
But then how can you find the zeros? Factor?
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