Solution Bank Polynomials Target F
Answers to the Practice Problems are below in random order.
\(\begin{align}&f(x)=3x^3-6x^2-3x+6\ & \ \ \ \ &f(x)=-\frac{1}{5}x^3-\frac{9}{5}x^2-\frac{23}{5}x-3\ & \ \ \ \ &f(x)=3x^3-6x^2-3x+6\\\\
&f(x)=\frac{1}{2}(x-4)^2(x+5)\ & \ \ \ \ &x=-2, 5, \pm2i\ & \ \ \ \ &y=0.0083x^2-0.4018x^2+6.5517x+3.8146\\\\
&f(x)=-2x^3+4x^2+10x-12\ & \ \ \ \ &f(x)=\frac{1}{3}x^3-\frac{2}{3}x^2-\frac{11}{3}x+4\ & \ \ \ \ &f(x)=-2(x+1)(x+3)(x-3)^2\\\\
&f(x)=x^3-3x^2-7x+21\ & \ \ \ \ &f(x)=(2x+1)(x-7)(x+3)\ & \ \ \ \ &f(x)=x^3-13x^2+17x+195\\\\
&f(x)=x^3-14x^2+69x-90\ & \ \ \ \ &f(x)=\frac{1}{4}x^4-2x^2+4\ & \ \ \ \ &y=40.1752\\\\
&f(x)=x^3-9x^2+4x-36\ & \ \ \ \ &f(x)=x^3+2x^2+25x+50\end{align}\)
\(\begin{align}&\text{Degree: 5}\\
&\text{Leading Coefficient: -2}\\
&\text{End Behavior: As} \:x\ \longrightarrow\ -\infty,\ f\left(x\right)\ \longrightarrow\ +\infty,\ \text{as}\ \: x\ \longrightarrow\ +\infty,\ f\left(x\right)\ \longrightarrow\ -\infty\\
&\text{x-intercepts: (-1,0), multiplicity 1, (2,0) and (4,0) both have multiplicity two}\\
&\text{y-intercpet: (0,-32)}\\
&\text{Domain:}\ (-\infty,\infty)\ \text{Range:} (-\infty,\infty)\end{align}\)
Answers to the Practice Problems are below in random order.
\(\begin{align}&f(x)=3x^3-6x^2-3x+6\ & \ \ \ \ &f(x)=-\frac{1}{5}x^3-\frac{9}{5}x^2-\frac{23}{5}x-3\ & \ \ \ \ &f(x)=3x^3-6x^2-3x+6\\\\
&f(x)=\frac{1}{2}(x-4)^2(x+5)\ & \ \ \ \ &x=-2, 5, \pm2i\ & \ \ \ \ &y=0.0083x^2-0.4018x^2+6.5517x+3.8146\\\\
&f(x)=-2x^3+4x^2+10x-12\ & \ \ \ \ &f(x)=\frac{1}{3}x^3-\frac{2}{3}x^2-\frac{11}{3}x+4\ & \ \ \ \ &f(x)=-2(x+1)(x+3)(x-3)^2\\\\
&f(x)=x^3-3x^2-7x+21\ & \ \ \ \ &f(x)=(2x+1)(x-7)(x+3)\ & \ \ \ \ &f(x)=x^3-13x^2+17x+195\\\\
&f(x)=x^3-14x^2+69x-90\ & \ \ \ \ &f(x)=\frac{1}{4}x^4-2x^2+4\ & \ \ \ \ &y=40.1752\\\\
&f(x)=x^3-9x^2+4x-36\ & \ \ \ \ &f(x)=x^3+2x^2+25x+50\end{align}\)
\(\begin{align}&\text{Degree: 5}\\
&\text{Leading Coefficient: -2}\\
&\text{End Behavior: As} \:x\ \longrightarrow\ -\infty,\ f\left(x\right)\ \longrightarrow\ +\infty,\ \text{as}\ \: x\ \longrightarrow\ +\infty,\ f\left(x\right)\ \longrightarrow\ -\infty\\
&\text{x-intercepts: (-1,0), multiplicity 1, (2,0) and (4,0) both have multiplicity two}\\
&\text{y-intercpet: (0,-32)}\\
&\text{Domain:}\ (-\infty,\infty)\ \text{Range:} (-\infty,\infty)\end{align}\)