Find the gradient of the line passing through the point (2,8) and touching the curve `y=x^3.`

Options

A. 3
B. 6
C. 9
D. 12

Answer

To find the gradient of the line passing through the point (2, 8) and touching the curve \( y = x^3 \), we can follow these steps:

### Step 1: Understand the problem
We need to find the slope of the tangent line to the curve \( y = x^3 \) at the point where it touches the line passing through the point (2, 8).

### Step 2: Differentiate the curve
To find the slope of the tangent line to the curve, we need to differentiate the function \( y = x^3 \) with respect to \( x \).

\[ \frac{dy}{dx} = 3x^2 \]

### Step 3: Evaluate the derivative at the point of tangency
We need to evaluate the derivative at the point \( x = 2 \) to find the slope of the tangent line at that point.

\[ \frac{dy}{dx} \bigg|_{x=2} = 3(2^2) = 3 \cdot 4 = 12 \]

### Step 4: Conclusion
The gradient (slope) of the line passing through the point (2, 8) and touching the curve \( y = x^3 \) is \( 12 \).

### Final Answer
The required gradient of the line is \( 12 \).

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