Calc 1: Find the point on the graph of f(x)=x^3 such that the tangent line at that point has an x-intercept of 6

Answer 1

We do not know the x-coordinate of where the tangent line touches the function f(x) = x³. Let's say that this point of tangency occurs at x = a. Since this point of tangency is on the graph of the function, the y-coordinate there must be f(a) = a³. Thus the point of tangency is at (a, a³). We also know that the tangent line goes through the point (6,0). Since we know two points on the tangent line, we can find the slope of the tangent line:

m = Δy/Δx = (a³ - 0) / (a - 6) = a³ / (a - 6)

However, since we are dealing with a tangent line we can also find its slope using a derivative. The derivative of f(x) = x³ is f '(x) = 3x². Thus the slope of the tangent line to the graph of f(x) at x = a must be f '(a) = 3a². We have now found the slope of the same tangent line using two different methods, once using Δy/Δx and then using a derivative. These two slopes must be the same because it is the same line. We can thus set the two expressions for the slope equal to each other and solve for a:

a³ / (a - 6) = 3a² → a = 3(a - 6) → 2a = 18 → a = 9

Thus the point on the graph where the line is tangent is (a, a³) = (9, 729)