Find the equation of the secant line given that x = 1 and x = 2. f(x) = x^2 - 4 Evaluate f(x) at x = 1 and x = 2. f(1) = (1)^2 - 4 f(1) = 1 - 4 f(1) = -3 One point is (1, -3). f(2) = (2)^2 - 4 f(2) = 4 - 4 f(2) = 0 Another point is (2, 0). Use (1, -3) to form a linear equation. y = mx + b_1 -3 = m(1) + b_1 -3 = m + b_1 -3 - m = b_1 Use (2, 0) to form another linear equation. y = mx + b_2 0 = m(2) + b_2 0 = 2m + b_2 0 - 2m = b_2 -2m = b_2 Now set b_1 = b_2 to find m. -m - 3 = -2m -3 = -2m + m - 3 = -m -3/-1 = m 3 = m I now have y = 3x + b. Use (1, -3) or (2, 0) to find b. I will use (2, 0). y = 3x + b 0 = 3(2) + b 0 = 6 + b 0 - 6 = m -6 = m. . .the slope of the secant line we seek. So, the equation of the secant line is y = 3x - 6. You say?
