`"Let "f(x) ={ underset( x^(2) +8 " "." "x ge0)(-x^(2)" "." "x lt0).` Equation of tangent line touching both branches of `y=f(x)` is

Options

A. `y= 4x+1`
B. `y=4x+4`
C. `y=x+4`
D. `y=x+1`

Answer

To find the equation of the tangent line touching both branches of the function \( f(x) \), we need to analyze the function and derive the equations for the tangent lines at the points of tangency.

### Step 1: Define the function
The function is given as:
\[ f(x) = \begin{cases} -x^2 & \text{if } x < 0 \\ x^2 + 8 & \text{if } x \geq 0 \end{cases} \]

### Step 2: Set up the equation of the tangent line
Assume the equation of the tangent line is:
\[ y = mx + c \]
where \( m \) is the slope and \( c \) is the y-intercept.

### Step 3: Tangent to the first branch \( f(x) = -x^2 \)
For the first branch \( y = -x^2 \) (where \( x < 0 \)), we equate the tangent line to the function:
\[ mx + c = -x^2 \]
Rearranging gives:
\[ x^2 + mx + c = 0 \]
For this quadratic equation to have exactly one solution (tangency), the discriminant must be zero:
\[ D = m^2 - 4c = 0 \quad \text{(1)} \]

### Step 4: Tangent to the second branch \( f(x) = x^2 + 8 \)
For the second branch \( y = x^2 + 8 \) (where \( x \geq 0 \)), we set up the equation:
\[ mx + c = x^2 + 8 \]
Rearranging gives:
\[ x^2 - mx + (8 - c) = 0 \]
Again, for tangency, the discriminant must be zero:
\[ D = (-m)^2 - 4(1)(8 - c) = 0 \]
This simplifies to:
\[ m^2 - 32 + 4c = 0 \quad \text{(2)} \]

### Step 5: Solve the system of equations
Now we have two equations:
1. \( m^2 - 4c = 0 \)
2. \( m^2 + 4c - 32 = 0 \)

From equation (1), we can express \( c \) in terms of \( m \):
\[ c = \frac{m^2}{4} \]
Substituting this into equation (2):
\[ m^2 + 4\left(\frac{m^2}{4}\right) - 32 = 0 \]
This simplifies to:
\[ m^2 + m^2 - 32 = 0 \implies 2m^2 - 32 = 0 \implies 2m^2 = 32 \implies m^2 = 16 \implies m = \pm 4 \]

### Step 6: Find \( c \)
Substituting \( m = 4 \) into \( c = \frac{m^2}{4} \):
\[ c = \frac{16}{4} = 4 \]
Substituting \( m = -4 \) into \( c = \frac{m^2}{4} \):
\[ c = \frac{16}{4} = 4 \]
Thus, in both cases, we have \( c = 4 \).

### Step 7: Write the equations of the tangent lines
The two possible equations of the tangent lines are:
1. \( y = 4x + 4 \) (for \( m = 4 \))
2. \( y = -4x + 4 \) (for \( m = -4 \))

### Conclusion
The equations of the tangent lines touching both branches of \( y = f(x) \) are:
1. \( y = 4x + 4 \)
2. \( y = -4x + 4 \)