Sketch the region bounded by the graphs of the equation \[ \begin{array}{l}f(x)=x^{2}+6 x \\ g(x)=x+6\end{array} \]

To sketch the region bounded by the graphs of the equations \( f(x) = x^2 + 6x \) and \( g(x) = x + 6 \), we first need to find the points of intersection. 1. Set \( f(x) = g(x) \): \[ x^2 + 6x = x + 6 \] Simplifying this, we get: \[ x^2 + 5x - 6 = 0 \] Factoring, we find: \[ (x + 6)(x - 1) = 0 \] This gives us the intersection points \( x = -6 \) and \( x = 1 \). 2. To identify the bounded region, we can evaluate the functions at a point between the intersections, say \( x = 0 \): \[ f(0) = 0^2 + 6 \cdot 0 = 0 \] \[ g(0) = 0 + 6 = 6 \] Since \( f(0) < g(0) \), the parabola \( f(x) \) lies below the line \( g(x) \) between \( x = -6 \) and \( x = 1 \). Now that we know \( f(x) \) is below \( g(x) \) in the interval \( [-6, 1] \), we can sketch the graphs: - The graph of \( f(x) = x^2 + 6x \) is a parabola that opens upwards, with its vertex located at \( x = -3 \). - The graph of \( g(x) = x + 6 \) is a straight line with a slope of 1. Place the intersection points at \( (-6, 0) \) and \( (1, 7) \) on the graph. The region bounded by these two curves will be closed off by the parabola from below and the line from above, forming a distinct shape on the graph between these x-values. Don't forget to shade the area between these two curves to represent the bounded region!