37. Which one is the point(s) intersection of graphs \( f \) and \( g \) when \( f(x)=-2 \) and \( g(x)=e^{x^{2}-x-2}-3 \) \( \begin{array}{ll}\text { (A) }(-1,-2) & \text { (B) }(2,0) \\ \text { (C) }(2,-2) & \text { (D) Both (A) }+ \text { (C) }\end{array} \)

To find the intersection points of the graphs \( f(x) = -2 \) and \( g(x) = e^{x^2 - x - 2} - 3 \), we set them equal to each other: \[ -2 = e^{x^2 - x - 2} - 3 \] Rearranging gives us \( e^{x^2 - x - 2} = 1 \). The exponential function is equal to 1 when its exponent equals zero, so we set: \[ x^2 - x - 2 = 0 \] Factoring the quadratic equation, we find \( (x - 2)(x + 1) = 0 \), giving us \( x = 2 \) and \( x = -1 \). We can substitute these back to find their corresponding \( y \)-values: - For \( x = -1 \): \( f(-1) = -2 \) (so point is \((-1, -2)\)). - For \( x = 2 \): \( f(2) = -2 \) (so point is \((2, -2)\)). Thus, the intersection points are both \((-1, -2)\) and \((2, -2)\), making the answer \( \text{(D) Both (A) + (C)} \). Now, here's some further information: Did you know that \( e^{x^2 - x - 2} \) can be interpreted as a transformation of the basic exponential function? This expression reflects how exponential functions grow based on the quadratic polynomial in their exponent, leading to fascinating behaviors in graphing! When solving equations, it’s common to overlook the importance of factoring, particularly with quadratics. Always double-check your work: for instance, ensure your factorization is correct and that you've considered all possible roots, as they might lead to points of intersection in multivariable equations or systems!