How do you find the vertex and intercepts for #y= 1/2x^2#?
This is trivial
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To find the vertex and intercepts for the equation ( y = \frac{1}{2}x^2 ):

Vertex: The vertex of a quadratic function in the form ( y = ax^2 + bx + c ) is at the point ((h, k)), where (h = \frac{b}{2a}) and (k) is the value of the function at (h).
For ( y = \frac{1}{2}x^2 ), (a = \frac{1}{2}), (b = 0). Substituting into the vertex formula: ( h = \frac{b}{2a} = \frac{0}{2 \cdot \frac{1}{2}} = 0 ) Evaluate ( y ) at ( h = 0 ): ( y = \frac{1}{2}(0)^2 = 0 )
So, the vertex is at ((0, 0)).

Intercepts:

xintercept: Set (y) to zero and solve for (x). (0 = \frac{1}{2}x^2 ) (x^2 = 0 ) (x = 0 )
So, the xintercept is at ((0, 0)).

yintercept: Set (x) to zero and solve for (y). (y = \frac{1}{2}(0)^2 = 0 )
So, the yintercept is also at ((0, 0)).

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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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