Final answer:
To find the equation of the tangent line to the curve y = -x² + 3x + 8 at x = 2, we need to find the slope of the curve at that point. The slope of the curve is found by taking the derivative of the function y with respect to x, which gives us dy/dx = -2x + 3. At x = 2, the slope is -1. Therefore, the equation of the tangent line is y = -x + 10.
Step-by-step explanation:
To determine the equation of the tangent line to the curve y = -x² + 3x + 8 at the point where x = 2, we need to find the slope of the curve at that point.
We can do this by finding the derivative of the function y with respect to x, which gives us the rate of change of y with respect to x. The derivative of y = -x² + 3x + 8 is dy/dx = -2x + 3.
At x = 2, the slope of the curve is -2(2) + 3 = -1. Therefore, the equation of the tangent line is y = mx + b, where m is the slope (-1) and b is the y-intercept.
Plugging in x = 2 and y = -x² + 3x + 8 into the equation, we can solve for b. y = -1(2) + b, y = b - 2. Since the point (2, y) lies on the tangent line, we substitute x = 2 into the original equation to solve for y. y = -2² + 3(2) + 8, y = -4 + 6 + 8, y = 10. Therefore, the equation of the tangent line is y = -x + 10.