Question

ons and use your results to sketch their graphs. b) $g(x)=-2x^{3}+4x^{2}+16x$

Explanation:

Step1: Find the x - intercepts

Set $g(x)=0$. So, $-2x^{3}+4x^{2}+16x = 0$. Factor out $-2x$: $-2x(x^{2}-2x - 8)=0$. Then factor the quadratic: $-2x(x - 4)(x+2)=0$. The x - intercepts are $x = 0$, $x = 4$ and $x=-2$.

Step2: Find the y - intercept

Set $x = 0$ in $g(x)$. Then $g(0)=-2(0)^{3}+4(0)^{2}+16(0)=0$.

Step3: Find the first - derivative

$g'(x)=-6x^{2}+8x + 16$. Factor out $-2$: $g'(x)=-2(3x^{2}-4x - 8)$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $3x^{2}-4x - 8 = 0$ where $a = 3$, $b=-4$, $c=-8$, we have $x=\frac{4\pm\sqrt{(-4)^{2}-4\times3\times(-8)}}{2\times3}=\frac{4\pm\sqrt{16 + 96}}{6}=\frac{4\pm\sqrt{112}}{6}=\frac{4\pm4\sqrt{7}}{6}=\frac{2\pm2\sqrt{7}}{3}$.

Step4: Analyze the end - behavior

As $x\to+\infty$, $y=-2x^{3}+4x^{2}+16x\to-\infty$ since the leading term $-2x^{3}$ dominates. As $x\to-\infty$, $y=-2x^{3}+4x^{2}+16x\to+\infty$.

Answer:

To sketch the graph: Plot the x - intercepts at $x=-2,0,4$ and y - intercept at $(0,0)$. Use the critical points from the first - derivative to find local maxima and minima. Consider the end - behavior: the graph goes up as $x\to-\infty$ and down as $x\to+\infty$.