How to minimize the distance between the point $(-1,0)$ and the curve $y^2 - x^3 = 0$? Given the equation of  a curve : $y^2 - x^3 = 0$, it is demanded to find the shortest distance from the point $(-1,0)$ to this curve.
Here is what I tried:

*

*solve the curve formula for $y$:
y²=x³

*write the function of the distance between$(-1,0)$ and the curve:
d(x,y)=√((1+x)²+y²)

*replaced in d $y^2$ with $x^3$ and squared $d(x,y)$:
x³+x²+2x+1=h(x)

*attempted to find the minimum of $h(x)$:
h'(x)=3x²+2x+2
but this derivate never equals to zero, so couldn't find any minimum for the function $h(x)$ because it tends -∞ when $x$ tends to -∞

Then i tried with Lagrange multipliers and found:

*

*$2y(1+λ)=0  \implies  λ=-1$ or $y=0$
I rejected $y=0$ because for y=0 gradient of $y^2-x^3$ isn't linearly independent which is one of the conditions to use Langrange method.

*$3x^2+2x+2=0$ (same as $h'(x)$ so it doesn't have a solution)
So I'm stuck with is and don't know how to solve the question, any help ?

 A: Since this is tagged "multivariable calculus", you have presumably dealt with parametric curves.  We can get around the problem created by the cusp in this "semicubical parabola" by parametrizing it using $ \ x(t) \ = \ t^2 \ , \ y(t) \ = \ t^3 \ \ . \ $  The distance-squared function for a point on the curve to $ \ (-1 \ , \ 0) \ $ is then $ \ (x + 1)^2 + y^2 \ \rightarrow \ (t^2 + 1)^2 + t^6 \ \ , \ $ which is an even function with a single evident absolute minimum.
If the point instead should be $ \ (1 \ , \ 0) \ \ , \ $ the distance-squared function becomes $ \ (t^2 - 1)^2 + t^6 \ \ , \ $ which is again even and has a relative maximum at $ \ t \ = \ 0 \ $ and two (symmetric) absolute minima.  Both functions have derivatives which are a factor of $ \ t \ $ times a biquadratic function of $ \ t \ \ , \ $ making extremization reasonably simple.
ADDENDUM (2/18) --  For this point, the approach, mentioned in the comments, using normal lines to the curve also works:  at a point $ \ ( x \ , \ x^{3/2} ) \ $ on the curve, the line to $ \ (1 \ , \ 0 ) \ $ has slope $ \ \large{\frac{x^{3/2}}{x - 1} } \ $ which must equal the slope of the normal line $ \ \large{ -\frac{2}{3·x^{1/2}} } \ \ . \ $  Setting these equal produces a quadratic equation for which only one value of $ \ x \ $ is positive; symmetry of the curve about the $ \ x-$axis then gives the two points closest to $ \ (1 \ , \ 0) \ \ . \  $   This also shows why the Lagrange method was not helpful for $ \ ( -1 \ , \ 0) \ \ : \ $ because of the character of its concavity, the normal lines to the curve cannot pass through that point, indicated by the analogous quadratic equation $ \ 3x^2 + 2x + 2 \ = \ 0 \ $ having no real solutions.
