1. **State the problem:** We need to sketch the function $$y=(x-4)(x^2+9)$$. 2. **Understand the function:** This is a product of two factors: a linear factor $x-4$ and a quadratic factor $x^2+9$. 3. **Important notes:** The quadratic $x^2+9$ is always positive since $x^2 \geq 0$ and adding 9 keeps it positive for all real $x$. 4. **Find the roots:** Set $y=0$: $$ (x-4)(x^2+9)=0 $$ Since $x^2+9 \neq 0$ for real $x$, the only root is from $x-4=0$, so: $$ x=4 $$ 5. **Determine the sign of $y$:** - For $x<4$, $x-4<0$ and $x^2+9>0$, so $y<0$. - For $x>4$, $x-4>0$ and $x^2+9>0$, so $y>0$. 6. **Find the y-intercept:** Set $x=0$: $$ y=(0-4)(0^2+9) = (-4)(9) = -36 $$ 7. **Find critical points (extrema):** Differentiate: $$ y = (x-4)(x^2+9) = x^3 - 4x^2 + 9x - 36 $$ $$ y' = 3x^2 - 8x + 9 $$ Set $y'=0$: $$ 3x^2 - 8x + 9 = 0 $$ Discriminant: $$ \Delta = (-8)^2 - 4 \cdot 3 \cdot 9 = 64 - 108 = -44 < 0 $$ No real roots for $y'$, so no critical points, meaning no local maxima or minima. 8. **Behavior at infinity:** As $x \to \pm \infty$, $y \approx x^3$ dominates, so $y \to \infty$ as $x \to \infty$ and $y \to -\infty$ as $x \to -\infty$. 9. **Summary:** - Root at $x=4$. - $y$ negative for $x<4$, positive for $x>4$. - No local extrema. - Y-intercept at $(0,-36)$. - Function behaves like cubic with positive leading coefficient. This information allows sketching the curve accurately.