Calculate the gradient of the tangent to the curve \( y=4 x^{2}-2 x+1 \) at \( (2,13) \). The curve \( y=a x^{2}-\frac{b}{x} \) has gradients 3 and -2 at \( x=1 \) and \( x=\frac{1}{2} \) respectively. Find the values of \( a \) and \( b \). \( \begin{array}{ll}\text { a) } \frac{d y}{d x}=8 x-2\end{array} \)

Let's address each part of your question step by step. ### 1. Gradient of the Tangent to the Curve \( y = 4x^2 - 2x + 1 \) at \( (2, 13) \) To find the gradient of the tangent to the curve at a specific point, we'll first find the derivative of the function, which represents the gradient function. Given: \[ y = 4x^2 - 2x + 1 \] **Step 1: Find the derivative \( \frac{dy}{dx} \).** \[ \frac{dy}{dx} = \frac{d}{dx}(4x^2) - \frac{d}{dx}(2x) + \frac{d}{dx}(1) = 8x - 2 + 0 = 8x - 2 \] **Step 2: Evaluate the derivative at \( x = 2 \).** \[ \left.\frac{dy}{dx}\right|_{x=2} = 8(2) - 2 = 16 - 2 = 14 \] **Conclusion:** The gradient of the tangent to the curve at the point \( (2, 13) \) is **14**. --- ### 2. Determining the Values of \( a \) and \( b \) for the Curve \( y = ax^2 - \frac{b}{x} \) Given the curve: \[ y = ax^2 - \frac{b}{x} \] We know the gradients at specific points: - Gradient is **3** at \( x = 1 \). - Gradient is **-2** at \( x = \frac{1}{2} \). **Step 1: Find the derivative \( \frac{dy}{dx} \).** \[ \frac{dy}{dx} = 2ax + \frac{b}{x^2} \] **Step 2: Set up equations based on the given gradients.** 1. At \( x = 1 \), \( \frac{dy}{dx} = 3 \): \[ 2a(1) + \frac{b}{1^2} = 3 \quad \Rightarrow \quad 2a + b = 3 \quad \text{(Equation 1)} \] 2. At \( x = \frac{1}{2} \), \( \frac{dy}{dx} = -2 \): \[ 2a\left(\frac{1}{2}\right) + \frac{b}{\left(\frac{1}{2}\right)^2} = -2 \quad \Rightarrow \quad a + 4b = -2 \quad \text{(Equation 2)} \] **Step 3: Solve the system of equations.** From **Equation 1**: \[ 2a + b = 3 \quad \Rightarrow \quad a = \frac{3 - b}{2} \] Substitute \( a \) from Equation 1 into **Equation 2**: \[ \frac{3 - b}{2} + 4b = -2 \] Multiply both sides by 2 to eliminate the fraction: \[ 3 - b + 8b = -4 \quad \Rightarrow \quad 3 + 7b = -4 \] \[ 7b = -7 \quad \Rightarrow \quad b = -1 \] Now, substitute \( b = -1 \) back into Equation 1 to find \( a \): \[ 2a + (-1) = 3 \quad \Rightarrow \quad 2a = 4 \quad \Rightarrow \quad a = 2 \] **Conclusion:** The values are: - \( a = 2 \) - \( b = -1 \) --- ### Summary 1. **Gradient at \( (2, 13) \):** \( \frac{dy}{dx} = 8x - 2 \). At \( x = 2 \), the gradient is **14**. 2. **Values of \( a \) and \( b \):** \( a = 2 \) and \( b = -1 \).