91Ó°ÊÓ

A cubic equation is a polynomial equation of the form \[ax^3 + bx^2 + cx + d = 0\]. In this exercise, we start with the cubic curve equation \[x^3 + y^3 - 12x - 8y - 16 = 0\]. To check if the curve touches the x-axis, we need to substitute \(y = 0\) into the equation. By doing so, the equation simplifies to \[x^3 - 12x - 16 = 0\]. This is now a cubic equation in one variable, \(x\). We solve this by finding the roots, starting with easy values. For example, testing \(x = 4\) shows that it satisfies the equation, making it a root. Thus, we can factorize the polynomial as: \[(x-4)(x^2 + 4x + 4) = 0\]. Solving the quadratic equation, \((x+2)^2 = 0\), confirms that \(x = 4\) is the only real root. These steps are essential for identifying where the cubic curve might touch the x-axis.

Implicit differentiation is a technique to find the derivative of an equation when it is not explicitly solved for one variable. For our exercise, we start with the curve \[x^3 + y^3 - 12x - 8y - 16 = 0\]. We differentiate both sides with respect to \(x\), remembering that \(y\) is also a function of \(x\): \[3x^2 + 3y^2 \frac{dy}{dx} - 12 - 8 \frac{dy}{dx} = 0\]. To isolate \(\frac{dy}{dx}\), we collect the terms involving \(\frac{dy}{dx}\) on one side: \[3x^2 - 12 = -3y^2 \frac{dy}{dx} - 8 \frac{dy}{dx}\]. We then factor out \(\frac{dy}{dx}\): \[3x^2 - 12 = -(3y^2 + 8) \frac{dy}{dx}\]. In our specific case, since we are looking for where the curve is horizontal (where the slope is zero), we set \(\frac{dy}{dx} = 0\) and solve for \(x\): \[3x^2 - 12 = 0\] which simplifies to \(x = 4\).

A point of tangency on a curve is where the curve just touches another line, in our case, the x-axis. For a curve to touch the x-axis, the y-coordinate of the point must be zero, and the slope of the tangent at that point must also be zero. Starting from our step-by-step exercise: we set \(y = 0\) in the equation \[x^3 + y^3 - 12 x - 8 y - 16 = 0\]. This transformation simplifies the expression to \(x^3 - 12x - 16 = 0\). The cubic equation has a real root at \(x = 4\). To confirm this point of tangency, we need further that the slope \(\frac{dy}{dx}\) at this point equals zero. From implicit differentiation, we have seen \(\frac{dy}{dx} = \frac{3x^2 - 12}{3y^2 + 8}\). Setting \(y = 0\) and substituting \(x = 4\), we showed that the derivative is \(\frac{dy}{dx} = 0\). Therefore, the curve indeed touches the x-axis at (4,0), confirming it as our point of tangency.