Question

Third-degree, with zeros of -4, -2, and 1, and a y-intercept of -11.

Answer 1

$\begin{cases} x = -4\implies &x+4=0\\ x=-2\implies &x+2=0\\ x=1\implies &x-1=0 \end{cases}\qquad \implies (x+4)(x+2)(x-1)=\stackrel{y}{0}$

now, that's the equation or polynomial in factored form, hmmm we also know that it has a y-intercept of -11, namely, when x = 0  y = -11, well let's plug in a factor to it, that will reflect those values, namely say hmmm factor "a", so

$(x+4)(x+2)(x-1)=y\qquad \stackrel{\textit{adding "a" factor for vertical shift}}{a(x+4)(x+2)(x-1)}=y \\\\\\ \stackrel{\textit{we know that when x = 0, y = -11}}{a(0+4)(0+2)(0-1)=-11}\implies -8a=-11\implies a=\cfrac{-11}{-8}\implies a = \cfrac{11}{8} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\mathbb{FOIL}}{\cfrac{11}{8}(x^2+6x+8)}(x-1)=y\implies \cfrac{11}{8}(x^3+6x^2+8x-x^2-6x-8)=y \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{11}{8}(x^3+5x^2+2x-8)=y~\hfill$