Final answer:
To find the equations of the tangent and normal lines to g(x) = 2−x at x = −7, first find the slope of the tangent, which is −1, and then use the point-slope form to find the tangent line: y = −x + 2. To find the normal line, use the negative reciprocal slope, 1, and the point-slope form to find y = x + 16.
Step-by-step explanation:
To find the equation of the tangent and normal line to the graph of g(x) = 2−x at x = −7, we first need to find the derivative of the function, which represents the slope of the tangent line. Since g(x) is a linear function, its derivative g'(x) is constant, and therefore, the slope is the coefficient of x which is −1. The value of the function at x = −7 is g( −7 ) = 2 − (−7) = 9. Thus, the point of tangency is (−7, 9).
To express the equation of the tangent line in slope-intercept form (y = mx + b), we can use the slope −1 and the point (−7, 9) to solve for b. Plugging the slope and point into the equation, we get 9 = (−1)(−7) + b, which simplifies to 9 = 7 + b; solving for b gives b = 2. Therefore, the equation of the tangent line is y = −x + 2.
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is −1, the slope of the normal line is 1. Using the point (−7, 9), and the slope 1, we plug into y = mx + b to find b. This gives us 9 = 1(−7) + b, simplifying to b = 16. Thus, the equation of the normal line is y = x + 16.