Rewriting f(x) as a Transformation of g(x)

  • Thread starter Thread starter halvizo1031
  • Start date Start date
  • Tags Tags
    Form Vertex
halvizo1031
Messages
77
Reaction score
0

Homework Statement



Can someone assist me with number 3 please...

Homework Equations





The Attempt at a Solution



I went ahead and rewrote f(x) in vertex form in terms of a,b, and c but I'm having a hard time writing down how f can be thought of as a transformation of g.
 

Attachments

  • scan0004.jpg
    scan0004.jpg
    67.8 KB · Views: 395
Physics news on Phys.org


What did you get? You have f(x) = ax2 + bx + c. Complete the square to get this into a form so that you can recognize the translations and other transformations relative to the graph of y = x2.
 


i got a[x+(b/2a)]^2 - [(b^2)/4a] + c where h=-(b/2a) and k=c-[(b^2)/4a]
 


halvizo1031 said:
i got a[x+(b/2a)]^2 - [(b^2)/4a] + c where h=-(b/2a) and k=c-[(b^2)/4a]
Shouldn't h be + b/(2a)?
So f(x) = a(x + b/(2a))2 + c - b2/(4a)

Or, f(x) = a(x + h)2 + k, with h = b/(2a) and k = c - b2/(4a).

You know what the graph of y = g(x) = x2 looks like, right? How would you need to transform the graph of g to get the graph of f?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top