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How do you find the minimum and maximum value of #y=3(x-3)^2-3#?

Answer 1

Minimum at #(3,-3)#

#y=a(x-h)^2+k# => Eqution of parabola in vertex form, where #(h, k)# is the vertex. For positive coefficient of squared term or a the parabola opens up, and if negative down. We have a minimum for upward and a maximum for down. So in this case: #y=3(x-3)^2-3# The parabola has a minimum value at vertex #(3, -3)#

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Answer 2

Jameson Allen

To find the minimum and maximum values of ( y = 3(x - 3)^2 - 3 ), we can first rewrite the equation in vertex form ( y = a(x - h)^2 + k ), where ( (h, k) ) represents the vertex of the parabola.

Given ( y = 3(x - 3)^2 - 3 ), we can identify ( h = 3 ) and ( k = -3 ). The coefficient ( a = 3 ), indicating that the parabola opens upwards.

Since the parabola opens upwards, the minimum value occurs at the vertex ( (h, k) ). Therefore, the minimum value of ( y ) is ( k = -3 ).

As for the maximum value, since the parabola opens upwards and does not have a maximum point (as it extends indefinitely upwards), there is no maximum value for ( y ) in this case.

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Answer from HIX Tutor

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.