Final answer:
The equation of the tangent line to the curve f(x) = x²+x that passes through the point (2, -3) and has a negative slope is y = -5x + 7.
Step-by-step explanation:
To find the equation of the tangent line to the curve, we need to find the derivative of the curve first. The derivative of f(x) = x²+x is f'(x) = 2x+1. Next, we need to find the slope of the tangent line at the point (2, -3). Plugging x = 2 into f'(x) gives us a slope of 5. Since we want a line with a negative slope, we need to negate the slope, giving us a slope of -5. Finally, using the point-slope form of a line, we can write the equation of the tangent line as y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope. Plugging in the values, we have y - (-3) = -5(x - 2). Simplifying, we get y + 3 = -5x + 10. Rearranging the equation in slope-intercept form gives us y = -5x + 7.