The tangent to the survey= `x^2 + 3x` will pass through the point (0, - 9) if jt is drawn at the point

Options

A. (3, 18)
B. (1, 4)
C. (-4,4)
D. (-3,0)

Answer

To solve the problem, we need to find the point on the curve \( y = x^2 + 3x \) where the tangent passes through the point \( (0, -9) \).

### Step-by-Step Solution:

1. **Find the derivative of the curve**:
The first step is to find the slope of the tangent line to the curve. The derivative of \( y \) with respect to \( x \) gives us the slope of the tangent.
\[ \frac{dy}{dx} = 2x + 3 \]

2. **Equation of the tangent line**:
The equation of the tangent line at a point \( (x_1, y_1) \) on the curve can be written using the point-slope form:
\[ y - y_1 = m(x - x_1) \]
where \( m = 2x_1 + 3 \) and \( y_1 = x_1^2 + 3x_1 \).

3. **Substituting the point (0, -9)**:
Since the tangent line must pass through the point \( (0, -9) \), we substitute \( x = 0 \) and \( y = -9 \) into the tangent line equation:
\[ -9 - (x_1^2 + 3x_1) = (2x_1 + 3)(0 - x_1) \]
Simplifying this gives:
\[ -9 - (x_1^2 + 3x_1) = -x_1(2x_1 + 3) \]
\[ -9 - x_1^2 - 3x_1 = -2x_1^2 - 3x_1 \]

4. **Rearranging the equation**:
Rearranging the equation leads to:
\[ -9 = -2x_1^2 - 3x_1 + x_1^2 + 3x_1 \]
\[ -9 = -x_1^2 \]
\[ x_1^2 = 9 \]

5. **Finding the values of \( x_1 \)**:
Taking the square root gives us:
\[ x_1 = 3 \quad \text{or} \quad x_1 = -3 \]

6. **Finding corresponding \( y_1 \) values**:
Now we need to find the corresponding \( y_1 \) values for both \( x_1 \):
- For \( x_1 = 3 \):
\[ y_1 = 3^2 + 3 \cdot 3 = 9 + 9 = 18 \]
- For \( x_1 = -3 \):
\[ y_1 = (-3)^2 + 3 \cdot (-3) = 9 - 9 = 0 \]

7. **Final points**:
Therefore, the points where the tangent touches the curve are:
\[ (3, 18) \quad \text{and} \quad (-3, 0) \]

### Conclusion:
The tangent to the curve \( y = x^2 + 3x \) will pass through the point \( (0, -9) \) if it is drawn at the points \( (3, 18) \) or \( (-3, 0) \).