#### The curve C has equation y= x^3 - 2x^2 - x +9,x 0 The point P has coordinates (2, 7).

Find the equation of the tangent to C at P, giving your answer in the form y = mx + c, where m and c are constants.

Anonymous

0

Krishna

0

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 that 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