Final answer:
To find the equation of the normal line to the given curve at x=1, first determine the slope of the tangent line using differentiation, then find the negative reciprocal (the slope of the normal line), use the point-slope form with the slope and given point, and finally express the equation in slope-intercept form.
Step-by-step explanation:
To find the equation of the normal line to the curve y = − 0.5 x^2 + 0.5x +3 at the point with abscissa x = 1, we must first determine the slope of the tangent line at that point. The slope of the tangent line is given by the derivative of the curve at x = 1. We then find the negative reciprocal of this slope to get the slope of the normal line. The equation of the straight line is typically given in the form y = mx + b, where m is the slope and b is the y-intercept.
The derivative of the given curve is dy/dx = -x + 0.5. Evaluating at x = 1 gives us the slope of the tangent line, which is -1 + 0.5 = -0.5. The slope of the normal line is then 2 since it is the negative reciprocal of -0.5. Now we know our normal line has a slope of 2 and passes through the point (1, 3), since substituting x = 1 in the curve equation yields y = 3. To find the y-intercept, we use the point-slope form of a straight line: y - y1 = m (x - x1), resulting in y - 3 = 2 (x - 1). Simplifying, we get the equation of the normal line as y = 2x + 1.