# How do you find the point on the the graph y=sqrtx which is plosest to the point (4,0)?

First note that the distance between an arbitrary point $\left(x , y\right) = \left(x , \sqrt{x}\right)$ on the graph of $y = \sqrt{x}$ and the point $\left(4 , 0\right)$ is $\sqrt{{\left(x - 4\right)}^{2} + {\left(\sqrt{x} - 0\right)}^{2}} = \sqrt{{x}^{2} - 7 x + 16}$.
Next, note that the value of $x$ where $d \left(x\right) = \sqrt{{x}^{2} - 7 x + 16}$ is minimized is the same value of $x$ where $s \left(x\right) = {\left(d \left(x\right)\right)}^{2} = {x}^{2} - 7 x + 16$ is minimized.
Since $s ' \left(x\right) = 2 x - 7$, $s \left(x\right)$ and $d \left(x\right)$ are minimized at $x = \frac{7}{2} = 3.5$ (note that $s ' ' \left(x\right) = 2 > 0$ so that the critical point is a minimum).
Therefore, the point on the graph of $y = \sqrt{x}$ that is closest to $\left(4 , 0\right)$ is $\left(\frac{7}{2} , \sqrt{\frac{7}{2}}\right) \setminus \approx \left(3.5 , 1.87083\right)$. The minimum distance is $d \left(\frac{7}{2}\right) = \sqrt{\frac{49}{4} - \frac{49}{2} + 16} = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2} \setminus \approx 1.93649$