Question

8-79. consider the function $y = x^3 - 9x$. homework help a. what are the zeros of the function? (factoring will help!) b. sketch a graph of the function.

Explanation:

Step1: Factor out common term

$y = x(x^2 - 9)$

Step2: Factor difference of squares

$y = x(x-3)(x+3)$

Step3: Set $y=0$ to find zeros

$x(x-3)(x+3) = 0$

Step4: Solve for $x$

$x=0$, $x-3=0 \implies x=3$, $x+3=0 \implies x=-3$

For part b:

Step1: Identify key features

Zeros at $x=-3,0,3$; leading term $x^3$ means end behavior: as $x\to+\infty$, $y\to+\infty$; as $x\to-\infty$, $y\to-\infty$

Step2: Find critical points

Take derivative: $y' = 3x^2 - 9$. Set to 0: $3x^2-9=0 \implies x^2=3 \implies x=\pm\sqrt{3}\approx\pm1.732$

Step3: Find extrema values

At $x=\sqrt{3}$: $y=(\sqrt{3})^3 -9\sqrt{3}=3\sqrt{3}-9\sqrt{3}=-6\sqrt{3}\approx-10.392$ (local min)
At $x=-\sqrt{3}$: $y=(-\sqrt{3})^3 -9(-\sqrt{3})=-3\sqrt{3}+9\sqrt{3}=6\sqrt{3}\approx10.392$ (local max)

Step4: Plot points and sketch

Plot zeros, local max/min, then draw a smooth cubic curve matching end behavior and passing through all points.

Answer:

a. The zeros of the function are $x=-3$, $x=0$, and $x=3$.
b. A sketch of the function will:

• Cross the x-axis at $(-3,0)$, $(0,0)$, and $(3,0)$
• Have a local maximum at $(-\sqrt{3}, 6\sqrt{3})$ and a local minimum at $(\sqrt{3}, -6\sqrt{3})$
• Increase to $+\infty$ as $x$ approaches $+\infty$, and decrease to $-\infty$ as $x$ approaches $-\infty$, with a smooth cubic curve connecting all key points.