The angle between x-axis and tangent of the curve `y=(x+1) (x-3)` at the point (3,0) is

Options

A. `tan^(-1).((8)/(15))`
B. ` tan^(-1).((15)/(8))`
C. `tan^(-1) 4`
D. none of these

Answer

To find the angle between the x-axis and the tangent of the curve \( y = (x + 1)(x - 3) \) at the point \( (3, 0) \), we can follow these steps:

### Step 1: Rewrite the equation of the curve
First, we can expand the equation of the curve:
\[ y = (x + 1)(x - 3) = x^2 - 3x + x - 3 = x^2 - 2x - 3 \]

### Step 2: Find the derivative \( \frac{dy}{dx} \)
Next, we need to find the derivative of \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(x^2 - 2x - 3) = 2x - 2 \]

### Step 3: Evaluate the derivative at the point \( (3, 0) \)
Now we will substitute \( x = 3 \) into the derivative to find the slope of the tangent line at that point:
\[ \frac{dy}{dx} \bigg|_{x=3} = 2(3) - 2 = 6 - 2 = 4 \]

### Step 4: Relate the derivative to the angle \( \theta \)
The slope of the tangent line is equal to \( \tan(\theta) \), where \( \theta \) is the angle between the tangent and the x-axis. Therefore:
\[ \tan(\theta) = 4 \]

### Step 5: Find the angle \( \theta \)
To find \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}(4) \]

### Conclusion
Thus, the angle between the x-axis and the tangent of the curve at the point \( (3, 0) \) is:
\[ \theta = \tan^{-1}(4) \]