The curve `y = ax ^(3) + bx ^(2) + cx + 5` touches the x-axis at `P (-2, 0)` and cuts the y-axis at a point Q where its gradient is 3, then `2a + 4b ` is equal to :

Options

A. `-5`
B. `-4`
C. `-3`
D. None of these

Explanation

To solve the problem step by step, we need to analyze the given information about the curve \( y = ax^3 + bx^2 + cx + 5 \).

### Step 1: Use the point where the curve touches the x-axis
Since the curve touches the x-axis at point \( P(-2, 0) \), we know that:
1. The point lies on the curve, so we can substitute \( x = -2 \) and \( y = 0 \) into the equation.
2. The derivative at this point must be zero since the curve touches the x-axis (it does not cross it).

Substituting \( x = -2 \) into the curve:
\[ 0 = a(-2)^3 + b(-2)^2 + c(-2) + 5 \]
This simplifies to:
\[ 0 = -8a + 4b - 2c + 5 \]
Rearranging gives us our first equation:
\[ 8a - 4b + 2c = 5 \quad \text{(Equation 1)} \]

### Step 2: Find the derivative and set it to zero
The derivative of the curve is:
\[ \frac{dy}{dx} = 3ax^2 + 2bx + c \]
Now, substituting \( x = -2 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=-2} = 3a(-2)^2 + 2b(-2) + c = 0 \]
This simplifies to:
\[ \frac{dy}{dx} \bigg|_{x=-2} = 12a - 4b + c = 0 \]
Rearranging gives us our second equation:
\[ 12a - 4b + c = 0 \quad \text{(Equation 2)} \]

### Step 3: Find the y-intercept and gradient at that point
The curve cuts the y-axis at point \( Q \) where \( x = 0 \):
\[ y = a(0)^3 + b(0)^2 + c(0) + 5 = 5 \]
Thus, the coordinates of point \( Q \) are \( (0, 5) \).

Given that the gradient at point \( Q \) is 3, we substitute \( x = 0 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=0} = c = 3 \]
So, we have:
\[ c = 3 \quad \text{(Equation 3)} \]

### Step 4: Substitute \( c \) into the equations
Now we substitute \( c = 3 \) into Equations 1 and 2.

**Substituting into Equation 1:**
\[ 8a - 4b + 2(3) = 5 \]
This simplifies to:
\[ 8a - 4b + 6 = 5 \implies 8a - 4b = -1 \quad \text{(Equation 4)} \]

**Substituting into Equation 2:**
\[ 12a - 4b + 3 = 0 \]
This simplifies to:
\[ 12a - 4b = -3 \quad \text{(Equation 5)} \]

### Step 5: Solve the system of equations
Now we have a system of two equations:
1. \( 8a - 4b = -1 \) (Equation 4)
2. \( 12a - 4b = -3 \) (Equation 5)

We can eliminate \( b \) by subtracting Equation 4 from Equation 5:
\[ (12a - 4b) - (8a - 4b) = -3 - (-1) \]
This simplifies to:
\[ 4a = -2 \implies a = -\frac{1}{2} \]

Now substitute \( a \) back into Equation 4 to find \( b \):
\[ 8(-\frac{1}{2}) - 4b = -1 \]
This simplifies to:
\[ -4 - 4b = -1 \implies -4b = 3 \implies b = -\frac{3}{4} \]

### Step 6: Calculate \( 2a + 4b \)
Now we can find \( 2a + 4b \):
\[ 2a + 4b = 2(-\frac{1}{2}) + 4(-\frac{3}{4}) = -1 - 3 = -4 \]

Thus, the final answer is:
\[ \boxed{-4} \]