Sketch the region bounded by the graphs of the equations. Sketch the region bounded by the graphs of the equations. Find the area of the region. \( \left.\qquad \begin{array}{r}f(x)=x^{2}+6 x \\ g(x)\end{array}\right)=x+6 \)

To sketch the region bounded by the curves \( f(x) = x^2 + 6x \) and \( g(x) = x + 6 \), first, we need to find their points of intersection. Setting \( f(x) = g(x) \): \[ x^2 + 6x = x + 6 \implies x^2 + 5x - 6 = 0 \] Factoring yields \((x - 1)(x + 6) = 0\), so the intersection points are \( x = 1 \) and \( x = -6 \). The area between the curves from \( x = -6 \) to \( x = 1 \) can be found by integrating the difference between the two functions: \[ \text{Area} = \int_{-6}^{1} [(x + 6) - (x^2 + 6x)]\,dx = \int_{-6}^{1} (-x^2 - 5x + 6)\,dx \] Calculating the integral: \[ \int (-x^2 - 5x + 6)\,dx = -\frac{x^3}{3} - \frac{5x^2}{2} + 6x \] Evaluating from \(-6\) to \(1\): \[ \left[-\frac{(1)^3}{3} - \frac{5(1)^2}{2} + 6(1)\right] - \left[-\frac{(-6)^3}{3} - \frac{5(-6)^2}{2} + 6(-6)\right] \] This results in: \[ \left[-\frac{1}{3} - \frac{5}{2} + 6\right] - \left[72 - 90 - 36\right] = \left[-\frac{1}{3} - \frac{15}{6} + 36/6\right] - \left[-54\right] \] Calculating yields an area of \( \frac{80}{3} \). Now, the area is approximately 26.67 square units. The sketch of this region will show a parabolic curve opening upward, dipping lower than a linear ascent of \( g(x) = x + 6 \), creating that lovely bounded section. If you want to dive a bit deeper into related functions, trying your hand at polynomial curve analysis is a blast! You can learn how to identify turning points and local maxima/minima, which is key in calculus. Moreover, if you're intrigued by real-world applications, consider how similar equations model projectile motion or trends in economics, illustrating theoretically how functions intersect and evolve over time. Exploring those could unleash your innovative side!