What is the domain, range, y intercepts and absolute maximum of #f(x)=8+2x-x^2#?

Question

2625 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-06T09:11:19

Domain #(-oo, +oo)#, Range #(-oo, +9)#, y-intercept #+8#, Maximum #+9#

#f(x) = 8+2x-x^2#

#f(x)# is a quadratic function, defined and continious #forall x in RR#

Hence, the domain of #f(x)# is #(-oo, +oo)#

The #y# intercept occurs at #x=0#

#-> y=8+0+0 = 8#

We know that the quadratic has an absloute maximum since the coefficient of #x^2 <0#

#f'(x) = 2-2x#

The absoulte maximum of #f(x)# occurs at #f'(x) = 0#

#-> 2x=2#

i.e #x=1#

#:. f(x)_"max" = 8+2-1 = 9#

Since #f(x)# has no finite minimum the range of #f(x)# is #(-oo, 9)#

These results can be realised by the graph of #f(x)# below:

graph{8+2x-x^2 [-15.97, 16.07, -4.59, 11.42]}