SOLUTION: Let the graph of g be a vertical stretch by a factor of 3 and a reflection in the y-axis, followed by a translation 2 units left of the graph of f(x)=x^2−6x+1. Write an equation.

Algebra ->  Linear-equations -> SOLUTION: Let the graph of g be a vertical stretch by a factor of 3 and a reflection in the y-axis, followed by a translation 2 units left of the graph of f(x)=x^2−6x+1. Write an equation.      Log On


   

Question 1147853: Let the graph of g be a vertical stretch by a factor of 3 and a reflection in the y-axis, followed by a translation 2 units left of the graph of f(x)=x^2−6x+1. Write an equation.
Found 2 solutions by CubeyThePenguin, ikleyn:
Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
g(x) = 3f(x+2)

= 3((x+2)^2 - 6(x+2) + 1)

= 3(x^2 + 4x + 4 - 6x - 12 + 1)

= 3(x^2 - 2x -7)

= 3x^2 - 6x - 21

Answer by ikleyn(52866) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The solution by  @CubeyThePenguin is  INCORRECT.

            I came to bring a correct solution. 


(1)  Apply vertical stretch by a factor 3.  You get then the polynomial

        p(x) = 3*(x^2 - 6x + 1).    (1)



(2)  Reflection in the y-axis is the  change  of x to (-x).

     So, after reflection, the new polynomial is 

         q(x) = 3*((-x)^2 - 6*(-x) + 1) = 3*(x^2 + 6x + 1).     (2)



(3)  Translation 2 units left is the change of x by (x+2) in the polynomial  (2).  

     So, the new polynomial g(x)   (your final answer)  is

         g(x) = 3*((x+2)^2 + 6(x+2) + 1) = 

                3*(x^2 + 4x + 4 + 6x + 12 + 1) = 3*(x^2 + 10x + 17) = 3x^2 + 30x + 51.


ANSWER.  The resulting polynomial is  g(x) = 3x^2 + 30x + 51.

Solved.

For your safety, ignore the post by  @CubeyThePenguin,
since it contains very serious errors.