For the function f, of, x, equals, 12, x, squared, plus, 2, x, plus, 5f(x)=12x 2 +2x+5, find the equation of the tangent line at x, equals, 5x=5.

To find the equation of the tangent line to the function f(x) = 12x^2 + 2x + 5 at x = 5, we need to find the slope of the tangent line (which is the derivative of the function at that point) and the y-coordinate of the point of tangency (which is the function value at that point).

Step 1: Find the derivative of the function. The derivative of f(x) = 12x^2 + 2x + 5 is f'(x) = 24x + 2.

Step 2: Evaluate the derivative at x = 5. f'(5) = 24*5 + 2 = 122. This is the slope of the tangent line.

Step 3: Find the y-coordinate of the point of tangency. f(5) = 125^2 + 25 + 5 = 325.

Step 4: Use the point-slope form of the equation of a line to find the equation of the tangent line. The point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Here, m = 122, x1 = 5, and y1 = 325, so the equation of the tangent line is y - 325 = 122(x - 5).