# Find the turning points of the following functions of x Y=2X2+7x+6/3X+4PLEASE COULD YOU WORK IT STEP BY STEPS? THANKS

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Remember what is the property that transforms any point into a turning point: if the derivative of the function changes the sign at a point, the point becomes a turning point.

Let's find the derivative of the function `y=(2x^2+7x+6)/(3x+4) ` using the quotient rule:

`dy/dx = ((2x^2+7x+6)'(3x+4)-(2x^2+7x+6)(3x+4)')/(3x+4)^2`

`dy/dx =((4x+7)(3x+4)-(2x^2+7x+6)(3)')/(3x+4)^2`

Open the brackets:

`dy/dx =(12x^2+37x+28-6x^2-21x-18)/(3x+4)^2`

`dy/dx =(6x^2+16x+10)/(3x+4)^2`

`dy/dx =2(3x^2+8x+5)/(3x+4)^2`

Check if the derivative is cancelling.

`2(3x^2+8x+5)/(3x+4)^2 = 0 lt=gt 3x^2+8x+5 = 0`

Use quadratic formula:

`x_(1,2) = (-8+-sqrt(64-60))/6`

`x_(1,2) =(-8+-2)/6 =gt x_1 = -6/6 = -1`

`=gt x_2 = -10/6 = -5/3`

The derivative changes its sign over the following intervals `(-oo,-5/3)` and `(-5/3,-1); (-5/3,-1)` and `(-1 ; +oo), ` therefore you may calculate the y coordinates of turning points at`x = -5/3` and `x = -1` .

`x = -5/3=gty=(50/9 - 35/3 + 6)/(-15/3 + 4) = (50 -105 + 54)/(-45+36) = 1/9`

`x = -1 =gt y=(2-7+6)/(-3+4) = 1`

**The turning points of the function y=(2x^2+7x+6)/(3x+4) are (-5/3 ; 1/9) and (-1 ; 1).**