Find an equation of the line tangent to the given curve at a.
y = 2x² / (3x - 1); a = 1

Verified step by step guidance

1. Step 1: Identify the function and the point of tangency. The function given is \( y = \frac{2x^2}{3x - 1} \) and the point of tangency is at \( a = 1 \).
2. Step 2: Find the derivative of the function to determine the slope of the tangent line. Use the quotient rule: if \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^2} \). Here, \( u = 2x^2 \) and \( v = 3x - 1 \).
3. Step 3: Calculate \( u' \) and \( v' \). For \( u = 2x^2 \), \( u' = 4x \). For \( v = 3x - 1 \), \( v' = 3 \).
4. Step 4: Substitute \( u, u', v, \) and \( v' \) into the quotient rule formula to find \( y' \). Simplify the expression to get the derivative.
5. Step 5: Evaluate the derivative at \( x = 1 \) to find the slope of the tangent line at the point \( a = 1 \). Use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point on the curve.