Sketch the region bounded by the graphs of the equations. Find the area of the region. \[ \begin{array}{l}y=x^{2}-4 \\ y=-x+3 \\ x=0 \\ x=2\end{array} \]

To sketch the region bounded by the equations \( y = x^2 - 4 \), \( y = -x + 3 \), \( x = 0 \), and \( x = 2 \), we'll start by identifying the points where these curves intersect within the specified bounds. 1. **Finding Points of Intersection**: To find the intersection of \( y = x^2 - 4 \) and \( y = -x + 3 \), we set them equal to each other: \[ x^2 - 4 = -x + 3 \] Rearranging gives: \[ x^2 + x - 7 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-1 \pm \sqrt{1 + 28}}{2} = \frac{-1 \pm \sqrt{29}}{2} \] Approximating gives us one root at roughly \( x \approx 1.19 \) (valid in our bounds) and another root at roughly \( x \approx -1.19 \) (which is outside the specified region). So, we compute the intersection point: \[ \text{At } x = \frac{-1+\sqrt{29}}{2}, \, y = -\left(\frac{-1+\sqrt{29}}{2}\right) + 3. \] After calculating, we find that the point of intersection \( \left(\frac{-1+\sqrt{29}}{2}, y\right) = \left(1.19, 1.81\right) \). 2. **Identifying the Bounded Area**: The equations bound the area we want between \( x = 0 \) and \( x = 2 \). At \( x = 0 \): \[ y = 0^2 - 4 = -4 \quad \text{and} \quad y = -0 + 3 = 3 \] Thus, the points are: \( (0, -4) \) and \( (0, 3) \). At \( x = 2 \): \[ y = 2^2 - 4 = 0 \quad \text{and} \quad y = -2 + 3 = 1 \] Thus, the points are: \( (2, 0) \) and \( (2, 1) \). 3. **Calculating the Area**: The area \( A \) can be found using the integral of the top function minus the bottom function from the left to the right bound (from \( x = 0 \) to \( x = 2 \)): \[ A = \int_0^2 \left[(-x + 3) - (x^2 - 4)\right] \, dx \] Simplifying the integrand: \[ = \int_0^2 (-x + 3 - x^2 + 4) \, dx = \int_0^2 (-x^2 - x + 7) \, dx \] Evaluating: \[ A = \left[-\frac{x^3}{3} - \frac{x^2}{2} + 7x\right]_0^2 \] Substituting \( x = 2 \): \[ = -\frac{8}{3} - 2 + 14 = -\frac{8}{3} + 12 = \frac{28 - 8}{3} = \frac{20}{3} \] Substituting \( x = 0 \) yields 0. Thus, the area of the region bounded by the curves and lines is \( \frac{20}{3} \) square units.