find the net signed area between the graph of...

# Explanation: ## Step1: Recall the definite - integral formula for net - signed area The net - signed area \(A\) between the graph of \(y = f(x)\) and the \(x\) - axis over the interval \([a,b]\) is given by \(A=\int_{a}^{b}f(x)dx\). Here, \(a=-8\), \(b = 1\), and \(f(x)=-x - 4\). So, \(A=\int_{-8}^{1}(-x - 4)dx\). ## Step2: Use the integral rules We know that \(\int_{-8}^{1}(-x - 4)dx=-\int_{-8}^{1}xdx-\int_{-8}^{1}4dx\). The power rule for integration is \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n\neq - 1\)). For \(\int_{-8}^{1}xdx\), using the power rule with \(n = 1\), we have \(\int_{-8}^{1}xdx=\left[\frac{x^{2}}{2}\right]_{-8}^{1}=\frac{1^{2}}{2}-\frac{(-8)^{2}}{2}=\frac{1}{2}-\frac{64}{2}=-\frac{63}{2}\). For \(\int_{-8}^{1}4dx\), since \(\int kdx=kx + C\) (where \(k\) is a constant), then \(\int_{-8}^{1}4dx=4x\big|_{-8}^{1}=4\times(1-( - 8))=4\times9 = 36\). ## Step3: Calculate the net - signed area \(A=-\left(-\frac{63}{2}\right)-36=\frac{63}{2}-36=\frac{63 - 72}{2}=-\frac{9}{2}\). # Answer: \(-\frac{9}{2}\)