3. The equation of a curve is \( y=\frac{x+4}{5-x} \). Find the gradient of the curve at the point where (i) the curve crosses the \( x \)-axis, (ii) the curve crosses the \( y \)-axis.

To find the gradient of the curve at specific points, we first need to calculate the derivative of the function \( y = \frac{x+4}{5-x} \). First, let’s find the derivative \( \frac{dy}{dx} \). Using the quotient rule, where \( u = x + 4 \) and \( v = 5 - x \): \[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \] Calculating \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = -1 \): \[ \frac{dy}{dx} = \frac{(5-x)(1) - (x+4)(-1)}{(5-x)^2} = \frac{5 - x + x + 4}{(5-x)^2} = \frac{9}{(5-x)^2} \] Now we can find the gradient at the \( x \)-axis and \( y \)-axis crossings. (i) To find where the curve crosses the \( x \)-axis, set \( y = 0 \): \[ 0 = \frac{x+4}{5-x} \implies x + 4 = 0 \implies x = -4 \] Now, substitute \( x = -4 \) into the derivative: \[ \frac{dy}{dx} \bigg|_{x=-4} = \frac{9}{(5 - (-4))^2} = \frac{9}{9^2} = \frac{9}{81} = \frac{1}{9} \] So, the gradient at the point where the curve crosses the \( x \)-axis is \( \frac{1}{9} \). (ii) To find where the curve crosses the \( y \)-axis, set \( x = 0 \): \[ y = \frac{0 + 4}{5 - 0} = \frac{4}{5} \] Now, substitute \( x = 0 \) into the derivative: \[ \frac{dy}{dx} \bigg|_{x=0} = \frac{9}{(5-0)^2} = \frac{9}{25} \] So, the gradient at the point where the curve crosses the \( y \)-axis is \( \frac{9}{25} \).