the vertex of the parabola represented by f(x)=x^2-2x+6 has coordinates (1,5). Find the coordinates of the vertex of the parabola defined by g(x)=f(x+3). Explain how you arrived at your answer

Answers

Answer 1

Answer: $Plug \ in \ x+3 \ into \ f(x) \ to \ get \ g(x) \n\ g(x) = (x+3)^2-2(x+3)+6 \n\ g(x)= x^2+6x+9 -2x-6+6 \n\ g(x) = x^2+4x+9 \n\ -(b)/(2a) --\ \textgreater \ vertex \n\ \frac {-4}{2} \n\ x \ coordinate = -2, plug \ it \ back \ into \ g(x) \n\ (-2)^2+4(-2)+9 \n\ 4-8 + 9 \n\ y \ coordinate = 5 \n\ \boxed{(-2,5)}$

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Answers

That's 33/100 and can't be simplified. If you had said that the 3s keep going and never end, then it would have been 1/3. But as written, it's just 33/100.