What is the range of f(x)=-x^2+5x+6?

Answer 1

f(x)= - x^2+5x+6
= - (x^2 - 5x - 6)
= - (x^2 - 5x +25/4 - 25/4 - 6)
= - (x - 5/2)^2 + 49/4

(x - 5/2)^2 > or = 0 for all real x
=> - (x - 5/2)^2 < or = 0
=> - (x - 5/2)^2 + 49/4 < or = 49/4
=> f(x) < or = 49/4
=> Range of f(x) is ( - infinity ,49/4)

Answer 2

firstly by inspection, the -x^2 factor would increase faster than the linear 5x factor, so we would expect much of the range to be negative.
now factorise into linear factors
-(x+2)(x+3)
roots are x = -2, -3
axis of symmetry is along line x = -2.5
substitute this into the equation to find positive limit of range (top of parabola), call this y, then the negative limit will be -infinity
so your range will be
{ -infinity < f(x) <= y }
or (-infinity, y]]

Also, we could have used calculus to find the maximum, and found the top part of the range that way, which would have been quicker, but i didnt know if you were doing calculus yet.

Answer 3

f(x)=x^2+5x+6
=x^2+3x+2x+6
=x(x+3)+2(x+3)
=(x+2) (x=3)
x = -2 x = -3

Answer 4

this is a parabola open downwards with a vertex at -5/-2 = 5/2

ymax = (5/2)^2 + 5(5/2) + 6

range will be -inf. to ymax