Summary: Determine the derivative of the function at that point (thus measuring the slope of the tangent line), determine #y#-coordinate, and utilize point-slope form to find the equation.
The derivative of a function #f(x)# at some point #x=x_0# measures the slope of the line tangent to the curve at that point. Recall that for a polynomial function of the form #f(x) = c_1x^n + c_2x^(n-1) + c_3x^(n-2) + ...+ c_nx + c_(n+1)#, where #c_1, c_2,# etc, are the constant coefficients in each term, the derivative #f'(x) = n*c_1x^(n-1) + (n-1)*c_2x^(n-2) + ...+c_n#.
Thus, our general form of the derivative for #f(x) = y = 3x^2 + 2x +1# is #f'(x) = (dy)/(dx) = 6x +2#.
Plugging in our desired #x_0#, #x_0=-1#...
#f'(-1) = 6(-1) +2 = -6+2 = -4#.
Note this only tells us the slope of the line, not the equation for the line itself. However, since the line is tangent to the function at #x=-1#, by finding the #y#-coordinate we would possess a set of coordinates and a slope, allowing us to use point-slope form.
#f(-1) = y = 3(-1)^2 + 2(-1) +1 = 3 -2 +1 = 2#
Thus the slope of the line tangent to #y = 3x^2 + 2x +1# at #(-1,2)# is #m=-4#. Recall the set-up for point slope form:
#y - y_0 = m(x - x_0)#
Plugging in #x_0 = -1, y_0 = 2, m = -4#, we get...
#y - 2 = -4(x+1) => y = -4(x+1)+2 => y = -4x -2#
Thus, the equation for the tangent line to the curve #y=3x^2 +2x +1# at #(-1,2)# is #y = -4x -2#
