The point on the curve `y=2x^(2)`, where the slope of the tangent is 8, is :

Options

A. `(0, 2)`
B. `(0, 8)`
C. `(2, 8)`
D. `(8, 2)`

Answer

To find the point on the curve \( y = 2x^2 \) where the slope of the tangent is 8, we can follow these steps:

### Step 1: Differentiate the function
The slope of the tangent to the curve at any point is given by the derivative \( \frac{dy}{dx} \).

Given:
\[ y = 2x^2 \]

Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2) = 4x \]

### Step 2: Set the derivative equal to the given slope
We know that the slope of the tangent is 8. Therefore, we set the derivative equal to 8:
\[ 4x = 8 \]

### Step 3: Solve for \( x \)
To find \( x \), we divide both sides by 4:
\[ x = \frac{8}{4} = 2 \]

### Step 4: Find the corresponding \( y \) value
Now that we have \( x = 2 \), we substitute it back into the original equation to find \( y \):
\[ y = 2(2^2) = 2 \times 4 = 8 \]

### Step 5: Write the point
Thus, the point on the curve where the slope of the tangent is 8 is:
\[ (2, 8) \]

### Final Answer
The point on the curve \( y = 2x^2 \) where the slope of the tangent is 8 is \( (2, 8) \).

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