PLEASE HELP GUYSSS?FEMALES What is the end behavior of the function f(x) = x3 + 2x2 + 4x + 5?
IS that 2x2 supposed to be 2 times 2
And is the x3 supposed to be \[x^3\]
yea
there are f(x)=x^3 +2x^2 +4x +5
^^^^
Okay hold on
So the limit as x approaches infinity?
?
I believe it falls to the left and rises to the right.
and as it approaches negative infinity too, yeah
i just learned this I have no idea what im supopsed to do
Answer choices?
What happens to the values of that function as x approaches -infinity?
@legggoheatnation
Up on the left, up on the right Up on the left, down on the right Down on the left, up on the right Down on the left, down on the right
Down on the left, up on the right
y
Negative numbers cubed on the right decreasing exponentially
idk bout that tbh lol
Did you try using a graphing calculator or plotting points on a drawing?
no
whats tht
A graph...
thanks captain obvious
i mean like what explains it
for negative values of x f(x) will start negative then get closer to 5 when x = 0 so the graph will rise from the left then continue to rise as x gets more positive
You should say the following: since the polynomial is a third degree polynomial with a positive leading coefficient then the left side will continue going down and the right side direction will be up forever also for further information try to type ODD DEGREE POLYNOMIALS
|dw:1463866833503:dw|
- a very rough drawing lol
Hope that helps :)
hmm iTHInk i get it thanks man
HI!!
this is much easier than it looks
the degree is odd, and the leading coefficient is 1 the end behavior is the same for any polynomial of odd degree and positive leading coefficient, i.e .the end behavior is the same as \(y=x\) |dw:1463867722867:dw|
what would the answer be then?
misty's answer is the best one
\[f \prime \left( x \right)=3x^2+4x+4=3\left( x^2+\frac{ 4 }{ 3 }x \right)+4\] \[=3\left( x^2+\frac{ 4 }{ 3 }x+\frac{ 4 }{ 9 }-\frac{ 4 }{ 9 } \right)+4\] \[=3\left( x+\frac{ 2 }{ 3 } \right)^2-\frac{ 4 }{ 3 }+4\] \[=3\left( x+\frac{ 2 }{ 3 } \right)^2+\frac{ 8 }{ 3 }>\frac{ 8 }{ 3 }>0\] f(x) is always increasing.
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