The curve C has equation, y= 9 - 4x - \frac{8}{x}. The point P on C has x-coordinate equal to 2. (a) Show that the equation of the tangent to C at the point P is y = 1 – 2x.

Step 1: Calculate the slope of the curve
Skill 1: Note down the given curve equation and apply the differentiation on
both sides
NOTE: Multiple both sides of the function by \frac{dy}{dx}
Skill 2: Differentiate both sides of the function with respect
to x (any variable).
EXAMPLE: y = x^3 - 2x^2 - x + 9
[math] \frac{d}{dx} y = \frac{d}{dx}[x^3] - \frac{d}{dx} [2x^2] -\frac{d}{dx} x + \frac{d}{dx} 9][/math]
\frac{dy}{dx} = 3x^2 - 4x - 1
Skill 3: Plug in (x, y) values to find ( \frac{dy}{dx})
for any point.
Skill 4: To know the slope (gradient), apply the BODMAS rules to simplify
the equation.
Step 2: Use the slope of curve and a point on curve to find its y-intercept.
Step 3:Use the slope of line and the y-intercept of line to find the equation of the line.
EXAMPLE: Plug the slope m = -2 and the y-intercept b = -8 into the slope- intercept formula.
y = mx + b
y = -2x + -8
Step 4: Verify that the calculated equation and given equation are similar or not.