Final answer:
To find a vertical tangent for the equation x³ - y² + x² = 0, we perform implicit differentiation and solve for dy/dx. We then look for points where the denominator of the derivative is zero, which happens at y = 0. The original equation shows x = 0 is a solution; hence, the curve has a vertical tangent at x = 0.
Step-by-step explanation:
To find the vertical tangent line for the equation x³ - y² + x² = 0, we need to implicitly differentiate the equation with respect to x because the equation is not explicitly solved for y. Implicit differentiation will provide us the derivative dy/dx, which we can analyze to find where the slope is vertical (undefined).
First, let's differentiate the equation:
- Derivative of x³ with respect to x is 3x².
- Derivative of -y² with respect to x is -2y(dy/dx) since y is a function of x.
- Derivative of +x² with respect to x is 2x.
So the differentiated equation is 3x² - 2y(dy/dx) + 2x = 0. We then solve for dy/dx to find the slope at any point (x, y):
2y(dy/dx) = 3x² + 2x
dy/dx = (3x² + 2x) / (2y)
The slope will be vertical where the denominator equals zero, so we look for points where y = 0. Substituting y = 0 back into the original equation x³ - y² + x² = 0, we find that x = 0 is a valid solution, meaning the curve has a vertical tangent at x = 0.