by Guest » Sun Sep 06, 2026 2:28 am
Hi, it's Ben.
Two pieces on the graph.
Left piece — cubic. Flat crossing (inflection) at [tex]x=-4[/tex], so it has the form [tex]y=a(x+4)^3[/tex]. It falls left-to-right, and at [tex]x=-6[/tex] we get [tex]y=8[/tex]:
[tex]a(-6+4)^3=8\ \Rightarrow\ -8a=8\ \Rightarrow\ a=-1[/tex]
So [tex]y=-(x+4)^3[/tex].
Right piece — absolute value. Vertex [tex](1,-1)[/tex], slopes [tex]\pm 1[/tex]:
[tex]y=|x-1|-1[/tex]
Answer:
[tex]f(x)=\begin{cases}-(x+4)^3, & -6\le x<-2\\[4pt] |x-1|-1, & -2<x\le 9\end{cases}[/tex]