91Ó°ÊÓ

To sketch the function \(f(x) = -1 - \sqrt{x}\), we need to identify key features such as intercepts, the general shape of the curve, and any asymptotes. This function is a transformation of the parent function, \(y = \sqrt{x}\). The negative sign in front of \(\sqrt{x}\) reflects the graph over the \(x\)-axis, and the \(-1\) term shifts the graph downward by 1 unit. The function is only defined for all non-negative values of \(x.\) The graph will look something like this: (Here, you can refer the graph from any online graphing calculator by searching "graph of -1 - sqrt(x)")

To find the intersection points with the x-axis, we need to determine when the function \(f(x)\) equals 0. $$0 = -1 - \sqrt{x}$$ Solve this equation for \(x\) to find the intersection point: $$\sqrt{x} = -1$$ Since the square root of a number cannot be negative, there is no intersection point with the \(x\)-axis.

Since there is no intersection with the x-axis, the region we're finding is completely below the x-axis, but the integral will result in a negative value. Thus, we'll find the absolute value of the integral to find the area. The definite integral to calculate the area A of the region bounded below by the graph and above by the x-axis from \(x=a\) to \(x=b\) is given by: $$A = \int_a^b \! f(x) \, \mathrm{d}x$$ $$A = \int_0^9 \! (-1-\sqrt{x}) \, \mathrm{d}x$$ Now, integrate the function with respect to \(x\) and evaluate the result for the given limits: $$A= \Big[ -x-\frac{2}{3}\sqrt[3]{x^3} \Big]_0^9$$ $$A= \Big( -9 - \frac{2}{3} (9) \Big) - \Big( 0 - \frac{2}{3}(0) \Big)$$ $$A= -27$$ Since the value of the area A is negative, take the absolute value to find the bounded area: $$A=|-27|=27$$ The area of the region bounded below by the graph of \(f(x)=-1-\sqrt{x}\) and above by the \(x\)-axis from \(x=0\) to \(x=9\) is 27 square units.