A differentiable function `y= f(x)` satisfies `f'(x)=(f (x))^2+5` and `f (0)= 1`. Then the equation of tangent at the point where the curve crosses y-axis, is

Options

A. `x-y+1=0`
B. `x-2y+1=0`
C. `6x-y+1=0`
D. `x-2y-1=0`

Answer

To find the equation of the tangent line to the curve \( y = f(x) \) at the point where it crosses the y-axis, we follow these steps:

### Step 1: Identify the point where the curve crosses the y-axis
The curve crosses the y-axis when \( x = 0 \). We are given that \( f(0) = 1 \). Therefore, the point of intersection is:
\[ (0, 1) \]

### Step 2: Find the derivative at \( x = 0 \)
We know from the problem statement that:
\[ f'(x) = (f(x))^2 + 5 \]
We need to find \( f'(0) \). Substituting \( x = 0 \) into the derivative equation gives:
\[ f'(0) = (f(0))^2 + 5 \]
Since \( f(0) = 1 \):
\[ f'(0) = (1)^2 + 5 = 1 + 5 = 6 \]

### Step 3: Write the equation of the tangent line
The equation of the tangent line can be expressed using the point-slope form:
\[ y - y_1 = m(x - x_1) \]
Where \( (x_1, y_1) \) is the point of tangency and \( m \) is the slope. Here, \( (x_1, y_1) = (0, 1) \) and \( m = 6 \):
\[ y - 1 = 6(x - 0) \]
This simplifies to:
\[ y - 1 = 6x \]
Rearranging gives:
\[ y = 6x + 1 \]

### Step 4: Write the final equation in standard form
To express the equation in standard form, we can rearrange it:
\[ 6x - y + 1 = 0 \]

Thus, the equation of the tangent at the point where the curve crosses the y-axis is:
\[ \boxed{6x - y + 1 = 0} \]