Question

Describe the end behavior of each function using arrow notation.1. f(x) = -x^3 + 7x^2 - 112. f(x) = -x^3 + 2x + 53. f(x) = 6x^5 - 4x

Answer 1

`\begin{gathered} 1)\text{ }As\text{ x }\rightarrow\text{ }\infty,\text{ f(x)}\rightarrow-\text{ }\infty \\ As\text{ x }\rightarrow\text{ -}\infty,\text{ f(x)}\rightarrow\text{ }\infty \\ 2)\text{ }As\text{ x }\rightarrow\text{ }\infty,\text{ f(x)}\rightarrow-\text{ }\infty \\ As\text{ x }\rightarrow\text{ -}\infty,\text{ f(x)}\rightarrow\text{ }\infty \\ 3)\text{ }As\text{ x }\rightarrow\text{ }\infty,\text{ f(x) }\rightarrow\text{ }\infty \\ As\text{ x }\rightarrow\text{ -}\infty,\text{ f(x) }\rightarrow\text{ -}\infty \end{gathered}`

Step-by-step explanation:
`\begin{gathered} 1)\text{ f(x) = }-x^3+7x^2-11 \\ \text{The leading coefficient = -1} \\ \text{The leading coefficient is negative} \\ \text{The exponent of the leading coefficient is odd} \\ As\text{ x }\rightarrow\text{ }\infty,\text{ f(x)}\rightarrow-\text{ }\infty \\ As\text{ x }\rightarrow\text{ -}\infty,\text{ f(x)}\rightarrow\text{ }\infty \end{gathered}`
`\begin{gathered} 2)\text{ }f\mleft(x\mright)=-x^3+2x+5 \\ \text{The leading coefficient = -1} \\ \text{The leading coefficient is negative} \\ \text{The exponent of the leading coefficient is 3 (odd)} \\ As\text{ x }\rightarrow\text{ }\infty,\text{ f(x) }\rightarrow\text{ -}\infty \\ As\text{ x }\rightarrow\text{ -}\infty,\text{ f(x) }\rightarrow\text{ }\infty \end{gathered}`
`\begin{gathered} 3)\text{ }f\mleft(x\mright)=6x^5-4x \\ \text{The leading coefficient = 6} \\ \text{The leading coefficient is positive} \\ \text{The exponent of the leading }coefficient\text{ is 5 (odd)} \\ As\text{ x }\rightarrow\text{ }\infty,\text{ f(x) }\rightarrow\text{ }\infty \\ As\text{ x }\rightarrow\text{ -}\infty,\text{ f(x) }\rightarrow\text{ -}\infty \end{gathered}`