which table describes the behavior of the graph of $f(x)=2x^{3}-26x - 24$?\ninterval\trelation of graph to x…

Explanation:

Step1: Find the roots of the function

Set $f(x)=2x^{3}-26x - 24 = 0$. First, factor out a 2: $2(x^{3}-13x - 12)=0$, or $x^{3}-13x - 12 = 0$. By trial - and - error, we find that $x=-1$ is a root. Then, using polynomial long - division or synthetic division, we divide $x^{3}-13x - 12$ by $(x + 1)$ to get $x^{2}-x - 12$. Factoring $x^{2}-x - 12=(x + 3)(x - 4)$. So the roots of $f(x)$ are $x=-3,x=-1,x = 4$.

Step2: Test intervals

We have the intervals $(-\infty,-3),(-3,-1),(-1,4),(4,\infty)$. For $x<-3$, let $x=-4$. Then $f(-4)=2(-4)^{3}-26(-4)-24=2(-64)+104 - 24=-128 + 104-24=-48<0$, so the graph is below the $x$ - axis on $(-\infty,-3)$. For $-3<x<-1$, let $x=-2$. Then $f(-2)=2(-2)^{3}-26(-2)-24=2(-8)+52 - 24=-16 + 52-24 = 12>0$, so the graph is above the $x$ - axis on $(-3,-1)$. For $-1<x<4$, let $x = 0$. Then $f(0)=2(0)^{3}-26(0)-24=-24<0$, so the graph is below the $x$ - axis on $(-1,4)$. For $x>4$, let $x = 5$. Then $f(5)=2(5)^{3}-26(5)-24=2(125)-130 - 24=250-130 - 24 = 96>0$, so the graph is above the $x$ - axis on $(4,\infty)$.

Answer:

Interval: $(-\infty,-3)$, Relation of graph to $x$-axis: Below Interval: $(-3,-1)$, Relation of graph to $x$-axis: Above Interval: $(-1,4)$, Relation of graph to $x$-axis: Below Interval: $(4,\infty)$, Relation of graph to $x$-axis: Above