Minimize Distance from Point to Parabola (Optimization) | Calculus 1 Exercises

We solve a common type of optimization problem where we are asked to find the points on a parabola that are closest to a given point. Thus, we are minimizing the distance from a point to a curve. To do this we begin with the distance formula, representing the distance between an arbitrary point on our parabola and the given point (0,2). Then we will do some basic algebra and eventually find the critical points of our function, which we classify using the first derivative test. #Calculus1 #apcalculus
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[@williamvasquez4857]

Why is 4 minus the negative square root = to 4 minus the positive square root in your very last equation?

[@WrathofMath]

Thanks for watching! I'm not 100% sure what you're describing, I see when x=-sqrt(3/2) in the second to last equation we have y = 4 - (-sqrt(3/2))^2. Since (-sqrt(3/2)) is being squared, the negative goes away and so too does the square root.

[@williamvasquez4857]

So basically -sqrt(x) is the same as +sqrt(x).

[@WrathofMath]

@@williamvasquez4857 No, (-sqrt(x))^2 is the same as sqrt(x)^2. Squaring the negative makes it positive, so if a negative quantity is being squared, the negative can be disregarded.