Calculus, 2ADV C1 2023 HSC 14 v1

Find the equation of the tangent to the curve `y=x(3x+2)^2` at the point `(1,25)`. (3 marks)

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`y=55x-30`

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`y`	`=x(3x+2)^2`
`dy/dx`	`=6x(3x+2) + (3x+2)^2`
	`=(3x+2)(9x+2)`

`text{At}\ x=1\ \ =>\ \ dy/dx=(3(1)+2)(9(1)+2)=55`

`text{Find equation of line}\ \ m=55,\ text{through}\ (1,25)`

`y-y_1`	`=m(x-x_1)`
`y-25`	`=55(x-1)`
`y`	`=55x-30`