Find the equation of the tangent line to the parabola y=x^2+3x+2 at point P(1, 6).

In order to find the equation of the tangent line, first we have to find its slope. To do this, we take the first derivative of the function. In this particular case, we just need to apply the power rule (if y=x^n, dy/dx=nx^(n-1)) to each of the terms: y=x^2+3x+2 => dy/dx=2x+3 Having done that, in order to find the slope at the particular point we're looking at, we have to substitute for the value of x we are given, in this case x=1. If the slope of the tangent line at point P is m, m=2x1+3=5 Finally, in order to find the equation of the tangent line, we can use the straight line equation, y-y1=m(x-x1), where (x1, y1) are the coordinates of the point we're given. By substituting, we find: y-6=5(x-1) => y-6=5x-5 => y=5x+1 So, the equation of the tangent line is y=5x+1.