How do I go about sketching the graph of
f(x)=$$-5x^4+28x^3-33x^2+2x+8$$

Question

How do I go about sketching the graph of 
f(x)=$$-5x^4+28x^3-33x^2+2x+8$$

cwrw238 · Answer

do you know calculus?

anonymous · Answer

no.

cwrw238 · Answer

you can differentiate in order to find any turning points on the graph
f'(x) = -20x^3 + 84x^2 - 66x + 2
equate this to zero and solve for x
to get the coordinates of turning points

find intersect on y axis by putting x = 0 
this is the point (0,8)
this should give you enough information to draw a sketch

cwrw238 · Answer

oh - i've just seen your post
-it would be difficult to sketch this without using calculus
you would have to plot points 
you could also try solving the equation

cwrw238 · Answer

- actually - it would be more tedious than difficult as you would have to plot a lot of points
solving the equation f(x) = 0 would be tedious also
to solve f(x) = 0 you'd need to find 1 root by trial and error then do long division or synthetic division  to find the others
- or use a graphical calculator!!

cwrw238 · Answer

from the numbers in the equation  ( -5 and 8 - the first and last) +/-2, +/-4 or +/-8 could be a solution

f(2) = -5(2)^4 + 28(2)^3 - 33(2)^2 + 2(2) + 8  not = 0
you can try -2, +/- 4  +/- 8 next

see  what i mean by tedious?