question\nfind the net signed area between the graph of the function f(x) = |x + 3| - 4 and the x - axis…

Explanation:

Step1: Split the absolute - value function

The function (y = f(x)=|x + 3|-4) can be split into two cases. When (x+3\geq0) (i.e., (x\geq - 3)), (y=(x + 3)-4=x - 1); when (x+3<0) (i.e., (x<-3)), (y=-(x + 3)-4=-x - 7).

Step2: Calculate the integral over sub - intervals

We split the interval ([-9,3]) into two sub - intervals ([-9,-3]) and ([-3,3]). The net - signed area (A=\int_{-9}^{-3}(-x - 7)dx+\int_{-3}^{3}(x - 1)dx). First, calculate (\int_{-9}^{-3}(-x - 7)dx). Using the power rule (\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have (\int(-x - 7)dx=-\frac{x^{2}}{2}-7x+C). Evaluating the definite integral: (\left(-\frac{(-3)^{2}}{2}-7\times(-3)\right)-\left(-\frac{(-9)^{2}}{2}-7\times(-9)\right)) [ \begin{align*} &(-\frac{9}{2}+21)-(-\frac{81}{2}+63)\ =&-\frac{9}{2}+21+\frac{81}{2}-63\ =&\frac{-9 + 81}{2}+21-63\ =&\frac{72}{2}+21 - 63\ =&36+21-63\ =& - 6 \end{align*} ] Second, calculate (\int_{-3}^{3}(x - 1)dx). Using the power rule, (\int(x - 1)dx=\frac{x^{2}}{2}-x+C). Evaluating the definite integral: (\left(\frac{3^{2}}{2}-3\right)-\left(\frac{(-3)^{2}}{2}-(-3)\right)) [ \begin{align*} &(\frac{9}{2}-3)-(\frac{9}{2}+3)\ =&\frac{9}{2}-3-\frac{9}{2}-3\ =& - 6 \end{align*} ]

Step3: Sum the results of the definite integrals

(A=-6+( - 6)=-12)

Answer:

(-12)