Question

e. $f(x)=-2x^{3}+2x^{2}-x + 5$
function type:
end behavior: as $x\to-\infty,f(x)\to$
end behavior: as $x\to\infty,f(x)\to$
f. $f(x)=\log _{2}x$
function type:
end behavior: as $x\to-\infty,f(x)\to$
end behavior: as $x\to\infty,f(x)\to$

1. use the graphs to describe the end behavior of each function by completing the statements.

end behavior: as $x\to-\infty,f(x)\to$
end behavior: as $x\to\infty,f(x)\to$

Explanation:

Step1: Analyze the degree and leading coefficient of \(f(x)=-2x^{3}+2x^{2}-x + 5\)

The function \(f(x)=-2x^{3}+2x^{2}-x + 5\) is a polynomial function. The degree \(n = 3\) (odd) and the leading coefficient \(a=-2\) (negative).
For a polynomial function \(y = a x^{n}+...\), when \(n\) is odd and \(a<0\):
As \(x\to-\infty\), \(y = f(x)\to\infty\) (because \(y=-2(-\infty)^{3}+...=-2(-\infty)=\infty\)).
As \(x\to\infty\), \(y = f(x)\to-\infty\) (because \(y=-2(\infty)^{3}+...=-2(\infty)=-\infty\)).

Step2: Analyze the domain and properties of \(y = \log_{2}x\)

The function \(y=\log_{2}x\) is a logarithmic function. The domain of \(y = \log_{2}x\) is \(x>0\).
As \(x\to-\infty\), \(x\) is not in the domain of \(y=\log_{2}x\), so \(f(x)\) is undefined.
As \(x\to\infty\), since the base \(b = 2>1\), \(\log_{2}x\to\infty\) (using the property of logarithmic functions \(y=\log_{b}x\), when \(b > 1\) and \(x\to\infty\), \(y\to\infty\)).

Step3: Analyze the graph (part a)

Looking at the graph (assuming it is a polynomial - like graph, since it has a smooth curve and no breaks).
As \(x\to-\infty\), the graph goes up (\(f(x)\to\infty\)).
As \(x\to\infty\), the graph goes down (\(f(x)\to-\infty\))

Answer:

e.
Function type: Polynomial function
End behavior: As \(x\to-\infty\), \(f(x)\to\infty\)
End behavior: As \(x\to\infty\), \(f(x)\to-\infty\)

f.
Function type: Logarithmic function
End behavior: As \(x\to-\infty\), \(f(x)\) is undefined
End behavior: As \(x\to\infty\), \(f(x)\to\infty\)

a.
End behavior: As \(x\to-\infty\), \(f(x)\to\infty\)
End behavior: As \(x\to\infty\), \(f(x)\to-\infty\)