Question

which table describes the behavior of the graph of f(x) = 2x^3 - 26x - 24? interval relation of graph to x - axis (-∞, - 3) above (-3, - 1) below (-1, 4) above (4, ∞) below interval relation of graph to x - axis (-∞, - 3) below (-3, - 1) above (-1, 4) below (4, ∞) above interval relation of graph to x - axis (-∞, - 4) above (-4, 1) below (1, 3) above (3, ∞) below

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 $x=-3$ is a root. Using synthetic division or long - division, we divide $x^{3}-13x - 12$ by $x + 3$ to get $x^{2}-3x - 4=(x + 1)(x - 4)$. So the roots of $f(x)$ are $x=-3,x=-1,x = 4$.

Step2: Test intervals

Choose test points in the intervals $(-\infty,-3),(-3,-1),(-1,4),(4,\infty)$. For the interval $(-\infty,-3)$, 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 the interval $(-3,-1)$, 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 the interval $(-1,4)$, 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 the interval $(4,\infty)$, 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:

The table with the intervals $(-\infty,-3)$ (Below), $(-3,-1)$ (Above), $(-1,4)$ (Below), $(4,\infty)$ (Above) describes the behavior of the graph of $f(x)$.