# Which point on the parabola #y=x^2# is nearest to (1,0)?

Approximately

I'm just going to solve this by the first method that comes to me, rather than trying to use any special geometric properties of parabolas.

The minimum will occur at a zero of:

graph{2x^3+x-1 [-10, 10, -5, 5]}

Using Cardano's method, find

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To find the point on the parabola y = x^2 nearest to (1,0), we need to minimize the distance between this point and (1,0). The point on the parabola can be represented as (x, x^2). Using the distance formula, the distance between (1,0) and (x, x^2) is sqrt((x - 1)^2 + (x^2 - 0)^2). To minimize this distance, differentiate the expression with respect to x, set it equal to zero, and solve for x. This yields x = 1/2. Therefore, the point on the parabola nearest to (1,0) is (1/2, 1/4).

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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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