Determine the equation of the line tangent to the graph of f (x) = x3 – 4x – 7 at x = 2.

Step 1: Find the derivative of the function f(x) = x^3 - 4x - 7. The derivative of a function gives us the slope of the tangent line at any point.

The derivative of x^3 is 3x^2, the derivative of -4x is -4, and the derivative of a constant like -7 is 0. So, the derivative of f(x) is f'(x) = 3x^2 - 4.

Step 2: Substitute x = 2 into the derivative to find the slope of the tangent line at x = 2.

f'(2) = 3(2)^2 - 4 = 3*4 - 4 = 12 - 4 = 8.

So, the slope of the tangent line at x = 2 is 8.

Step 3: Find the y-coordinate of the point on the function where x = 2. We do this by substituting x = 2 into the original function.

f(2) = (2)^3 - 4*2 - 7 = 8 - 8 - 7 = -7.

So, the point on the function where x = 2 is (2, -7).

Step 4: Use the point-slope form 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.

Substituting our values, we get y - (-7) = 8(x - 2), which simplifies to y + 7 = 8x - 16.

Step 5: Solve for y to get the equation of the tangent line in slope-intercept form (y = mx + b).

y = 8x - 16 - 7 = 8x - 23.

So, the equation of the line tangent to the graph of f(x) = x^3 - 4x - 7 at x = 2 is y = 8x - 23.