Question

Find the equation of tangent to the curve \( y = 2x^3 + x^2 + 2 \) at \( \left( \frac{1}{2}, 2 \right) \).

Hint:

Tangent equation requires the derivative for slope and point-slope form: \( y - y_1 = m (x - x_1) \).

Answer 1

Step 1: Verify the point lies on the curve: 
\[ y = 2 \left( \frac{1}{2} \right)^3 + \left( \frac{1}{2} \right)^2 + 2 = 2 \cdot \frac{1}{8} + \frac{1}{4} + 2 = \frac{2}{8} + \frac{1}{4} + 2 = \frac{1}{4} + \frac{1}{4} + 2 = 2. \] Point \( \left( \frac{1}{2}, 2 \right) \) satisfies the curve. 
 

Step 2: Find the slope of the tangent (derivative at \( x = \frac{1}{2} \)): 
\[ \frac{dy}{dx} = \frac{d}{dx} (2x^3 + x^2 + 2) = 6x^2 + 2x. \] At \( x = \frac{1}{2} \): 
\[ \frac{dy}{dx} = 6 \left( \frac{1}{2} \right)^2 + 2 \cdot \frac{1}{2} = 6 \cdot \frac{1}{4} + 1 = \frac{3}{2} + 1 = \frac{5}{2}. \]

Step 3: Equation of tangent using point-slope form \( y - y_1 = m (x - x_1) \): 
\[ y - 2 = \frac{5}{2} \left( x - \frac{1}{2} \right) \Rightarrow y - 2 = \frac{5}{2} x - \frac{5}{4}. \] \[ y = \frac{5}{2} x - \frac{5}{4} + 2 = \frac{5}{2} x - \frac{5}{4} + \frac{8}{4} = \frac{5}{2} x + \frac{3}{4}. \] Answer: \( y = \frac{5}{2} x + \frac{3}{4} \).