Solve the following inequality and graph the solution. Answer using interval \( x^{2}+2 x-3 \geq 0 \)

To solve the inequality \( x^{2} + 2x - 3 \geq 0 \), we first find the roots of the corresponding equation \( x^{2} + 2x - 3 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 2 \), and \( c = -3 \). Plugging in these values, we calculate: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-3)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm \sqrt{16}}{2} = \frac{-2 \pm 4}{2} \] This gives us the roots \( x = 1 \) and \( x = -3 \). Next, we analyze the sign of the quadratic expression in the intervals determined by the roots: \( (-\infty, -3) \), \( (-3, 1) \), and \( (1, \infty) \). 1. In the interval \( (-\infty, -3) \), choose \( x = -4 \): \( (-4)^2 + 2(-4) - 3 = 16 - 8 - 3 = 5 \) which is positive. 2. In the interval \( (-3, 1) \), choose \( x = 0 \): \( 0^2 + 2(0) - 3 = -3 \) which is negative. 3. In the interval \( (1, \infty) \), choose \( x = 2 \): \( 2^2 + 2(2) - 3 = 4 + 4 - 3 = 5 \) which is positive. Now, putting this together, the solution to the inequality \( x^{2} + 2x - 3 \geq 0 \) means we include the intervals where the expression is positive or zero. Therefore, the solution is: \[ (-\infty, -3] \cup [1, \infty) \] For graphing, plot the points \( -3 \) and \( 1 \) on a number line. Shade the regions to the left of \( -3 \) (including \( -3 \)) and to the right of \( 1 \) (including \( 1 \)) to demonstrate the solution visually. Final answer in interval notation: \[ (-\infty, -3] \cup [1, \infty) \]