It is given that the equation of a straight line which passes through point \( (0,-4) \) is \( y=-\frac{1}{2} x+c \). Find the point of intersection of the straight line and the \( x \)-axis. A \( (-8,0) \) B \( (-2,0) \) C \( (2,0) \) D \( (8,0) \)

To find the point of intersection of the straight line \( y = -\frac{1}{2}x + c \) with the \( x \)-axis, we need to set \( y = 0 \). Plugging this into the line's equation gives us: \[ 0 = -\frac{1}{2}x + c \] Next, we know the line passes through the point \( (0, -4) \). This means when \( x = 0 \), \( y = -4 \), so: \[ -4 = -\frac{1}{2}(0) + c \] Thus, \( c = -4 \) and our line's equation is now: \[ y = -\frac{1}{2}x - 4 \] Setting \( y = 0 \): \[ 0 = -\frac{1}{2}x - 4 \] Adding 4 to both sides: \[ 4 = -\frac{1}{2}x \] Multiplying by -2: \[ x = -8 \] So the point of intersection with the \( x \)-axis is \( (-8, 0) \). Therefore, the answer is A \( (-8,0) \).