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ScanSolve AI Answer
Here's the solution to finding the equation of the tangent to the curve at the given point: The curve is given by the equation y = x^3 - 3x^2 + 2x - 1. The point of tangency is (3, 5). Step 1: Find the first derivative of the curve to determine the slope of the tangent line. y = x^3 - 3x^2 + 2x - 1 (dy)/(dx) = (d)/(dx)(x^3) - (d)/(dx)(3x^2) + (d)/(dx)(2x) - (d)/(dx)(1) (dy)/(dx) = 3x^2 - 3(2x) + 2(1) - 0 (dy)/(dx) = 3x^2 - 6x + 2 Step 2: Evaluate the derivative at the given x-coordinate of the point of tangency, x=3, to find the slope m. m = .(dy)/(dx)|_x=3 = 3(3)^2 - 6(3) + 2 m = 3(9) - 18 + 2 m = 27 - 18 + 2 m = 9 + 2 m = 11 Step 3: Use the point-slope form of a linear equation, y - y_1 = m(x - x_1), with the point (x_1, y_1) = (3, 5) and the slope m = 11. y - 5 = 11(x - 3) Step 4: Simplify the equation to the standard form y = mx + c. y - 5 = 11x - 33 y = 11x - 33 + 5 y = 11x - 28 The equation of the tangent to the curve at the point (3, 5) is y = 11x - 28. Last free one today — make it count tomorrow, or type /upgrade for unlimited.