How do you find the axis of symmetry and vertex point of the function: #y=1/20x^2#?

Answer 1

If there are only #x^2#'s involved, the answer is allways
#x=0and (0,0)#

A non-square #x# or a number in the function would displace both the axis and the vertex. The 1/20 only makes it flatter as compared to the standard #x^2#-graph. graph{0.05x^2 [-20.44, 20.1, -4.54, 15.74]}
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Answer 2

The axis of symmetry for a parabola in the form y = ax^2 is the vertical line passing through the vertex, given by x = -b/(2a) where a and b are coefficients of the quadratic equation. For the given function y = (1/20)x^2, the coefficient a = 1/20. Thus, the axis of symmetry is x = 0. The vertex point lies on this axis of symmetry. To find the vertex point, substitute the value of x = 0 into the original equation. So, y = (1/20)(0)^2 = 0. Therefore, the vertex point is (0, 0).

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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.

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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