If f(v) is expressed as d = v + .2v^2,
why is f(2v) expressed as d = 2v + .8v^2?
If all terms are doubled, wouldn't it be expressed as d = 2v + .4v^2? The textbook doesn't explain why the second coefficient is quadrupled rather than doubled.

Question

If f(v) is expressed as d = v + .2v^2, 
why is f(2v) expressed as d = 2v + .8v^2?  
If all terms are doubled, wouldn't it be expressed as d = 2v + .4v^2?  The textbook doesn't explain why the second coefficient is quadrupled rather than doubled.

anonymous · Answer

Remember that when something is written as f(2v), we're plugging that into all values of v.

anonymous · Answer

Thus, f(2v) = 2v + 0.2 (2v)^2

anonymous · Answer

f(2v) = 2v + (0.2)(4)(v^2)

anonymous · Answer

f(2v) = 2v + 0.8v^2

anonymous · Answer

I think you forgot to square the two.

anonymous · Answer

OH!  I was looking at it as (2v) + (.2) (2v^2)

anonymous · Answer

In my logic, I was thinking that if d = v + .2 v^2 then the entire expression doubled would be d = 2v + .4 (2v^2)

anonymous · Answer

Yes, it's an easy mistake to make. If the entire expression is doubled, then it would be written as 2f(v) instead of f(2v)- but I don't think you need to learn that yet.

anonymous · Answer

Darlin, I am 40 years old and am about to get my AAS in Cisco Network Admin.  That is EXACTLY what I needed to learn :)

anonymous · Answer

What? Oh my gosh, I assumed you were taking introductory algebra and were a year or two behind me. xD

anonymous · Answer

it simply has been 21 years since I was in high school.  This class is Tech Math aka College Algebra.  I simply need a few things reminded :)