Find the stationary point of y=3x^2-12x+29 and classify it as a maximum/minimum

The stationary point of a curve is where it changes direction, that is the gradient goes from positive to negative or vice versa. At the exact point of the change, the gradient of the curve will be exactly 0.

This means we need to differentiate the equation of the curve and find the point at which the derivative is equal to 0. Using the rule that ax^b differentiates to abx^(b-1) we get:

(dy/dx)=6x-12

Setting this equal to 0 we get:

6x-12=0

6x=12

x=2,

Subbing this value of x into the original equation we find the y-value to be 17

This means the stationary point is the point (2,17)

To classify this point as a maximum/minimum we must examine the second derivative. We know that if the second derivative is positive the point is a minimum and if it is negative the point is a maximum. 

Differentiating (dy/dx)=6x-12 we get:

(d^2y/dx^2)=6

As this is always positive, we can classify the stationary point (2,17) as a minimum.

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