# How do you graph f(x)=x^5+3x^2-x using zeros and end behavior?

Feb 22, 2018

$\text{First we search the zeros}$

${x}^{5} + 3 {x}^{2} - x = x \left({x}^{4} + 3 x - 1\right)$
${x}^{4} + 3 x - 1 = \left({x}^{2} + a x + b\right) \left({x}^{2} - a x + c\right)$
$\implies b + c - {a}^{2} = 0 ,$
$\text{ } a \left(c - b\right) = 3 ,$
$\text{ } b c = - 1$
$\implies b + c = {a}^{2} , \text{ } c - b = \frac{3}{a}$
$\implies 2 c = {a}^{2} + \frac{3}{a} , \text{ } 2 b = {a}^{2} - \frac{3}{a}$
$\implies 4 b c = {a}^{4} - \frac{9}{a} ^ 2 = - 4$
$\text{Name k = a²}$

$\text{Then we get the following cubic equation}$
${k}^{3} + 4 k - 9 = 0$
$\text{Substitute k = r p :}$
${r}^{3} {p}^{3} + 4 r p - 9 = 0$
$\implies {p}^{3} + \left(\frac{4}{r} ^ 2\right) p - \frac{9}{r} ^ 3 = 0$
$\text{Choose r so that 4/r² = 3 => r = } \frac{2}{\sqrt{3}}$
$\text{Then we get}$
$\implies {p}^{3} + 3 p - \left(\frac{27}{8}\right) \sqrt{3} = 0$
$\text{Substitute p = t - 1/t :}$
$\implies {t}^{3} - \frac{1}{t} ^ 3 - \left(\frac{27}{8}\right) \sqrt{3} = 0$
$\implies {t}^{6} - \left(\frac{27}{8}\right) \sqrt{3} {t}^{3} - 1 = 0$
$\text{Substitute u = t³, then we get a quadratic equation}$
$\implies {u}^{2} - \left(\frac{27}{8}\right) \sqrt{3} u - 1 = 0$
$\mathrm{di} s c : 3 \cdot {\left(\frac{27}{8}\right)}^{2} + 4 = \frac{2443}{64}$
$\implies u = \frac{\left(\frac{27}{8}\right) \sqrt{3} \pm \frac{\sqrt{2443}}{8}}{2}$
$\implies u = \frac{27 \sqrt{3} \pm \sqrt{2443}}{16}$

$\text{Take the solution with the + sign : }$
$u = 6.0120053$
$\implies t = 1.8183317$
$\implies p = 1.2683771$
$\implies k = 1.4645957$
$\implies a = 1.2102048$
$\implies b = - 0.50716177$
$\implies c = 1.9717575$

${x}^{4} + 3 x - 1 = \left({x}^{2} + a x + b\right) \left({x}^{2} - a x + c\right)$
$\text{So the roots are}$
$x = \frac{- a \pm \sqrt{{a}^{2} - 4 \cdot b}}{2}$
$\implies x = - 0.6051024 \pm 0.93451094$
$\implies x = - 1.53961334 \text{ OR } 0.32940854$
$\text{And}$
$x = \frac{a \pm \sqrt{{a}^{2} - 4 \cdot c}}{2}$
$\implies x = \text{not real as } {a}^{2} - 4 \cdot c < 0$

$\text{So we have three zeros for our original quintic equation :}$
$x = = - 1.53961334 \text{ OR " 0 " OR } 0.32940854$

$\text{The end behavior is}$
${\lim}_{x \to - \infty} = - \infty \text{ , and}$
lim_{x->+oo} = +oo."

$\text{So we have}$

$- \infty \text{........." -1.53961334 "........." 0 ".........." 0.32940854 "........} + \infty$
$- - - - - - 0 + + + + 0 - - - - 0 + + + + +$