Algebra - Function

Question

I keep redoing this problem and cannot see where I am messing up, I know I am not getting the correct answer because it is not even an option listed (multiple choice problem).

Given f(r) = 2r²-4r+3, evaluate f(1-t)+2

Here is what I keep doing:

2(1-t)2+2-4(1-t)+2+3

(1-t)(1-t)
1-t-t+t²
t²-2t+1

2(t²-2t+1)+2-4(1-t)+2+3
2t²-4t+2+2-4+4t+2+3
2t²+4-4+2+3
2t²+5

I know I must be doing something wrong, any help is greatly appreciated.

Thank you in advance.

Answer 1

You made the after expanding the square term.  It should be

 

2(1 - t)(1 - t) - 4(1 - t) + 4 + 2

 

2(1 - 2t + t²) - 4(1  - t) + 4 + 2

 

 

Continue from here.

Answer 2

Hi Katie,

You appear to be adding 2 twice. You should be only adding 2 once to f(1-t)

 

f(1-t) = 2(1-t)² - 4(1-t) + 3

 

f(1-t) = 2(t² - 2t +1) - 4 + 4t + 3

 

f(1-t) = 2t² - 4t + 2 - 4 + 4t + 3

 

f(1-t) = 2t² + 1

 

So f(1-t) + 2 = 2t² + 1 + 2

 

f(1-t) + 2 = 2t² + 3