Question

College Algebra (Multiplicity): So I was wondering... How would a graph f(x)=4(x^2+1)(x-2)^3 look like? Since x^2 has no real zeros...

Answer 1

what's its end behavior and why?

Answer 2

@ganeshie8 @Hero

Answer 3

easy, since the degree of polynomial is `5`, it goes in different directions at ends

Answer 4

and since the leading coefficient is positive the end behavior is `DOWN-UP `

Answer 5

So I would count the x^2 even though it doesn't have any real zeroes then?

Answer 6

for end behavior, zeroes are of no use

Answer 7

we only look at "degree" and "leading coefficient"

Answer 8

Right. Thank you! :) That makes so much sense.

Answer 9

odd degree => different directions
even degree => same directions