Take-home Assignment \#01 1. Find the slop of the lure and the equation of the tangent line to the function \[ f(x)=3 x^{2}-4 x+1 \quad \text { at point when } x=2 \]

Sure, let's go through the problem step by step. ### Problem Statement Find the **slope** of the **curve** and the **equation of the tangent line** to the function \[ f(x) = 3x^{2} - 4x + 1 \] **at the point where \( x = 2 \).** ### Step 1: Find the Slope of the Curve at \( x = 2 \) The **slope** of the curve at a particular point is given by the **derivative** of the function at that point. 1. **Find the derivative** \( f'(x) \) of the function \( f(x) = 3x^{2} - 4x + 1 \): \[ f'(x) = \frac{d}{dx}(3x^{2}) - \frac{d}{dx}(4x) + \frac{d}{dx}(1) = 6x - 4 \] 2. **Evaluate the derivative** at \( x = 2 \) to find the slope \( m \): \[ f'(2) = 6(2) - 4 = 12 - 4 = 8 \] So, the **slope** \( m \) of the curve at \( x = 2 \) is **8**. ### Step 2: Find the Equation of the Tangent Line at \( x = 2 \) The equation of the tangent line can be written in the point-slope form: \[ y - y_1 = m(x - x_1) \] where: - \( m \) is the slope of the tangent line, - \( (x_1, y_1) \) is the point of tangency. 1. **Find the point of tangency** \( (x_1, y_1) \): - We already know \( x_1 = 2 \). - Calculate \( y_1 = f(2) \): \[ f(2) = 3(2)^{2} - 4(2) + 1 = 3(4) - 8 + 1 = 12 - 8 + 1 = 5 \] So, \( y_1 = 5 \). The point of tangency is \( (2, 5) \). 2. **Write the equation of the tangent line** using the point-slope form: \[ y - 5 = 8(x - 2) \] Simplify it to the slope-intercept form \( y = mx + b \): \[ y - 5 = 8x - 16 \\ y = 8x - 16 + 5 \\ y = 8x - 11 \] So, the **equation of the tangent line** at \( x = 2 \) is: \[ y = 8x - 11 \] ### Summary - **Slope of the curve at \( x = 2 \):** \( m = 8 \) - **Equation of the tangent line at \( x = 2 \):** \( y = 8x - 11 \)