# How do you find the critical points for # y = x^3 + 12x^2 + 6x + 8 #?

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To find the critical points of ( y = x^3 + 12x^2 + 6x + 8 ), you need to first find the derivative of the function and then solve for points where the derivative equals zero.

The derivative of the function ( y ) with respect to ( x ) is ( y' = 3x^2 + 24x + 6 ).

Set ( y' ) equal to zero and solve for ( x ):

( 3x^2 + 24x + 6 = 0 )

Now, you can use methods like factoring, completing the square, or the quadratic formula to solve for ( x ).

In this case, you can use the quadratic formula:

( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} )

where ( a = 3 ), ( b = 24 ), and ( c = 6 ).

Substitute these values into the formula and solve for ( x ). After finding the values of ( x ), these are the critical points of the function.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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