# Transformation

• Jul 27th 2010, 05:52 PM
ASD2010
Transformation
Suppose for the moment that b^2-4ac >0, So that f(x) has two distinct real roots , and a>0.

A) Consider the transformation f(x)----> f(x)+k. What is the value of K such that f(x)+k has two identical real roots: Hint, you will need to complete squares.

B) What is the condition on f(x)=ax^2+bx+c such that f(x) -----> f(-x) leaves the function unchanged? Hint: draw a picture of a quadratic function that does not change under this transformation. What kind of transformation is this?

• Jul 27th 2010, 08:47 PM
bondesan
For A:

If $f(x)=ax^2 + bx + c$, then $f(x)+k = ax^2 + bx + (c+k)$.

Two identical roots: $\dfrac{\sqrt{b^2 - 4a(c+k)}}{2a} = 0$.

For B:

I don't know precisely what do you mean in "leaving the function unchanged", but I guess you are looking for a condition that makes f(x)=f(-x). So what do we need for $ax^2+bx+c=a(-x)^2 + b(-x) + c$?
• Jul 27th 2010, 09:30 PM
ASD2010
Thank You very much, that really helps :)