Find the shortest distance between point $(2,0)$ and function $\sqrt{16x^2 + 5x+16}$. I want to find the shortest distance between point $(2,0)$ and function $\sqrt{16x^2 + 5x+16}$, So I did the optimization but the answer is always wrong.
$$\begin{align}D=&\sqrt{(x-2)^2+(y)^2}\\ y^2=&{16x^2 + 5x+16}\\ &\text{(by substituting)}\\ D=&\sqrt{(x-2)^2+(16x^2 + 5x+16)}\\ D=& \sqrt{17{x}^{2}+ x + 20}\\ &\text{(Convert to a polynomial since root doesn't matter in the domain)}\\ D(x)=& 17x^2+x+20 \\ D^\prime(x)=& 34x+1 \\ x =& \frac{-1}{34}\end{align}$$
Which is clearly not the closest point, So what did I do wrong?
Here is the question and my answer, which is evaluated as wrong.

 A: Your result is correct. The confusion is caused by the software you use to vizualize the graph - they do scale the axis differently AND they hide values of $y$ (see some graphs below for omitted values between $0$ and $3$). 
Maple (even with 1:1 ratio checked, values hidden on $y$ axis):

Google: (all $y$ values shown, but scale is different)

Mathematica (with AspectRatio->1, again missing values on $y$ axis)

You need to be careful about this and setup tools properly. Here is for example Geogebra which is using 1:1 scale by default and does not hide any values from the plot since you see whole canvas (https://www.geogebra.org/graphing). You can see that your result now makes sense visually.

A: We can minimize $(x-2)^2+y^2$ for convenience, since squaring is increasing on positive reals. Then as you substituted
$$
f(x)=(x-2)^2+16x^2+5x+16\Rightarrow f(x)=17x^2+x+20\Rightarrow f'(x)=34x+1=0\Rightarrow x=-\frac{1}{34}
$$
As you found. Making the closest point on the curve 
$$
(-\frac{1}{34},\sqrt{16(1/34)^2-5/34+16})\sim (-.03,3.99)
$$
What's the problem?
